Solution:
Let the large square have side length $2$, and the inscribed circle has radius $1$ (since the diameter equals the side of the square).

Let $s$ be the side length of one of the small squares. Place the large square with vertices at $(0,0)$, $(2,0)$, $(2,2)$, $(0,2)$, and the circle centered at $(1,1)$.

Consider the small square in the lower left corner. Let its vertices be $A = (0,0)$, $B = (s,0)$, $C = (s,s)$, $D = (0,s)$. The side $CD$ is tangent to the circle, and $A$ and $B$ lie on the sides of the large square.

The equation of the circle is $(x-1)^2 + (y-1)^2 = 1$.

The line $y = s$ (side $CD$) is tangent to the circle. The distance from the center $(1,1)$ to the line $y = s$ is $|1 - s|$.

Set this equal to the radius $1$:
$$|1 - s| = 1 \implies s = 0 \text{ or } s = 2$$
But $s = 0$ is degenerate, and $s = 2$ is the whole square, so this is not correct. Let's reconsider.

The small square is rotated so that one side is tangent to the circle, and two vertices are on the sides of the large square. Let the side of the small square be $a$.

Let the small square in the lower left corner have one vertex at $(0,0)$, and its sides along the axes, so its other vertices are $(a,0)$, $(a,a)$, $(0,a)$. The side $y = a$ is tangent to the circle.

The distance from $(1,1)$ to $y = a$ is $|1 - a|$. For tangency, $|1 - a| = 1$, so $a = 0$ or $a = 2$.

But this would make the square degenerate or as large as the big square. This suggests the small squares are rotated $45^{\circ}$.

Let us suppose the small square is rotated $45^{\circ}$, with one vertex at $(0,0)$, and the two adjacent vertices on the axes.

Let the center of the small square be at $(h, h)$, and its side length $a$. The vertices are at $(h, h) \pm \left(\frac{a}{2}, 0\right)$ and $(h, h) \pm \left(0, \frac{a}{2}\right)$, but rotated $45^{\circ}$.

Let us instead let the side of the small square be $a$, and its sides are parallel to the axes, but its position is such that one side is tangent to the circle, and two vertices are on the sides of the large square.

Alternatively, consider the distance from the center $(1,1)$ to the corner $(0,0)$: $\sqrt{2}$. The distance from $(1,1)$ to the side $y = x$ is $|1 - 0|/\sqrt{2} = 0$.

Alternatively, let's try a geometric approach.

Let us draw the diagonal from $(0,0)$ to $(2,2)$. The center of the circle is at $(1,1)$. The distance from $(1,1)$ to the diagonal is $0$.

Let us try to find the side length $a$ of the small square such that its side is tangent to the circle, and two vertices are on the sides of the large square.

Let us try to find the area by another method.

Let $O$ be the center of the circle. The four small squares are congruent and occupy the four corners outside the circle but inside the large square.

The area outside the circle but inside the square is $4 - \pi$.

But this region is made up of the four small squares and the four segments between the squares.

But the four small squares are the largest squares that can fit in the corners, tangent to the circle.

Let us focus on one corner, say the lower left.

Let the small square have side $a$, with one vertex at $(0,0)$, and the other at $(a,0)$, $(a,a)$, $(0,a)$.

The side $y = a$ is tangent to the circle. The distance from $(1,1)$ to $y = a$ is $|1 - a|$.

Set $|1 - a| = 1$, so $a = 0$ or $a = 2$.

But this is not possible. Therefore, the small squares are rotated $45^{\circ}$.

Let us suppose the small square is centered at $(h, h)$, and its vertices are $(h \pm \frac{a}{2\sqrt{2}}, h \pm \frac{a}{2\sqrt{2}})$.

Let us suppose the side of the small square is $a$, and its center is at $(c, c)$.

The vertices are at $(c + \frac{a}{2\sqrt{2}}, c + \frac{a}{2\sqrt{2}})$, $(c - \frac{a}{2\sqrt{2}}, c + \frac{a}{2\sqrt{2}})$, $(c - \frac{a}{2\sqrt{2}}, c - \frac{a}{2\sqrt{2}})$, $(c + \frac{a}{2\sqrt{2}}, c - \frac{a}{2\sqrt{2}})$.

The two vertices on the axes are $(c - \frac{a}{2\sqrt{2}}, c + \frac{a}{2\sqrt{2}})$ and $(c + \frac{a}{2\sqrt{2}}, c - \frac{a}{2\sqrt{2}})$.

Set $c - \frac{a}{2\sqrt{2}} = 0$ and $c + \frac{a}{2\sqrt{2}} = a_1$.

But $c - \frac{a}{2\sqrt{2}} = 0 \implies c = \frac{a}{2\sqrt{2}}$.

So the center is at $\left(\frac{a}{2\sqrt{2}}, \frac{a}{2\sqrt{2}}\right)$.

The side of the square is $a$.

The side of the square is tangent to the circle. The distance from the center of the circle $(1,1)$ to the side of the small square is $d = 1$.

The equation of the side of the small square that is tangent to the circle is $x + y = a/\sqrt{2}$.

The distance from $(1,1)$ to the line $x + y = a/\sqrt{2}$ is:
$$\frac{|1 + 1 - a/\sqrt{2}|}{\sqrt{2}} = \frac{|2 - a/\sqrt{2}|}{\sqrt{2}}$$
Set this equal to $1$:
$$\frac{|2 - a/\sqrt{2}|}{\sqrt{2}} = 1$$
So $|2 - a/\sqrt{2}| = \sqrt{2}$.

So $2 - a/\sqrt{2} = \sqrt{2}$ or $2 - a/\sqrt{2} = -\sqrt{2}$.

First, $2 - a/\sqrt{2} = \sqrt{2}$:
$$2 - \sqrt{2} = a/\sqrt{2} \implies a = (2 - \sqrt{2})\sqrt{2} = 2\sqrt{2} - 2$$
Second, $2 - a/\sqrt{2} = -\sqrt{2}$:
$$2 + \sqrt{2} = a/\sqrt{2} \implies a = (2 + \sqrt{2})\sqrt{2} = 2\sqrt{2} + 2$$
But $a$ must be less than $2$, so $a = 2\sqrt{2} - 2$.

Therefore, the area of one small square is:
$$(2\sqrt{2} - 2)^2 = (2\sqrt{2})^2 - 2 \cdot 2\sqrt{2} \cdot 2 + 2^2 = 8 - 8\sqrt{2} + 4 = 12 - 8\sqrt{2}$$
So the total area of the four small squares is:
$$4 \times (12 - 8\sqrt{2}) = 48 - 32\sqrt{2}$$
But this is too large. Let's check the calculation:

$(2\sqrt{2} - 2)^2 = (2\sqrt{2})^2 - 2 \cdot 2\sqrt{2} \cdot 2 + 2^2 = 8 - 8\sqrt{2} + 4 = 12 - 8\sqrt{2}$

So the total area is $4 \times (12 - 8\sqrt{2}) = 48 - 32\sqrt{2}$.

But the side of the large square is $2$, so the total area is $4$.

Therefore, the area of one small square is $12 - 8\sqrt{2}$, which is $\approx 0.686$.

Therefore, the total area is $4 \times (3 - 2\sqrt{2}) = 12 - 8\sqrt{2}$.

Final Answer:

The total area of the four smaller squares is $\boxed{12 - 8\sqrt{2}}$.