We start with three circles of radius $1$ that are tangent to each other externally and consider all rectangles that contain these circles such that each side is tangent to at least one of the circles. A square that contains the three circles and has sides parallel to one such rectangle has side length not smaller than each of the sides of this rectangle. Therefore, we seek the rectangle whose largest side is the smallest possible among all such rectangles.

Because the rectangle has four sides and there are only three circles, there is one circle to which two sides of the rectangle are tangent. These two sides must be adjacent (not opposite) because the other two circles are not contained between any two parallel tangents to one of the three circles.

Let $A$ be the centre of the circle that touches two sides of the rectangle, at points $P$ and $Q$. The other two circles have centres $B$ and $C$ and they are in contact with a rectangle side at points $R$ and $S$, respectively. The common tangent of the circles centred at $A$ and $C$ touches circle centre $A$ at $T_1$. Similarly, $T_2$ is the point of contact on this circle of the common tangent of the circles centred at $A$ and $B$. Note that both, $P$ and $Q$, need to be on the smaller arc $T_1T_2$ for the rectangle to include all three circles. We set
$$\alpha = \angle T_1AP \quad \text{and} \quad \beta = \angle T_2AQ.$$
Because $\angle T_1AT_2 = 120^\circ$ and $\angle PAQ = 90^\circ$ we have $\alpha + \beta = 30^\circ$. After reflecting, if necessary, in the line through $A$ which is tangent to the other two circles, we can assume without loss of generality that $\alpha \le \beta$ and hence $0 \le \alpha \le 15^\circ$.

The point $M$ is chosen on the line $SC$ so that $\angle CMA$ is a right angle. Similarly, $N$ lies on $PA$ giving right angle $\angle BNA$. Since $PA, QA, RB, SC$ are radii of the circles, the side lengths of the rectangle are $a = 2 + |CM|$ and $b = 2 + |AN|$.
Because a tangent is perpendicular to the radius through the point of contact, and $AB$ is parallel to the tangent through $T_2$, and $AC$ is parallel to the tangent through $T_1$, the angle equalities indicated in the diagram follow. In particular, $\angle ACM = \alpha$ and $\angle NAB = \beta$. Since $|AB| = |AC| = 2$ we then obtain $|CM| = 2\cos(\alpha)$ and $|AN| = 2\cos(\beta)$. The side lengths of the rectangle are therefore
$$a = 2 + 2\cos(\alpha) \quad \text{and} \quad b = 2 + 2\cos(\beta).$$

Since $0 \le \alpha \le 15^\circ$ and $\beta \le 30^\circ$ we have $a \ge b$ with equality iff $\alpha = \beta = 15^\circ$. Moreover, the Cosine function is decreasing for angles between $0^\circ$ and $180^\circ$, hence the smallest possible value for $a$ is achieved when $\alpha = 15^\circ$, the largest possible value in our situation. In this case, the rectangle is a square with side length $a = 2 + 2\cos(15^\circ)$. To express this value more concretely, we recall the trigonometric identity
$$\cos(x - y) = \cos(x) \cos(y) + \sin(x) \sin(y)$$
which we apply to $x = 45^\circ$ and $y = 30^\circ$. Since $\cos(45^\circ) = \sin(45^\circ) = \sqrt{2}/2$, $\cos(30^\circ) = \sqrt{3}/2$ and $\sin(30^\circ) = 1/2$ we finally obtain
$$a = 2 + 2\cos(15^\circ) = 2 + \frac{\sqrt{6} + \sqrt{2}}{2}.$$

If it is known that the smallest side length is realised by the square which has all four sides tangent to the circles, its side length can be calculated as follows.
We set up a coordinate system as shown below, with centre of the left circle at $(\sqrt{2}, 0)$ and centre of the upper circle at $(\sqrt{2} + \sqrt{3}, 1)$. The fitting square has one vertex at $(0, 0)$ and $x$-axis reflection symmetry.

The side of the square which touches the upper circle is perpendicular to a radius of slope $1$ of this circle. Its point of contact, $P$, therefore has coordinates $(\sqrt{2} + \sqrt{3} + \sqrt{2}/2, 1 + \sqrt{2}/2)$ and the equation of this tangent is
$$x + y = \sqrt{2} + \sqrt{3} + \sqrt{2}/2 + 1 + \sqrt{2}/2 = 2\sqrt{2} + \sqrt{3} + 1.$$

The $y$-intercept of this line is at $(0, 2\sqrt{2} + \sqrt{3} + 1)$. The length $d$ of the diagonal of the square is therefore equal to $2\sqrt{2} + \sqrt{3} + 1$, hence the side length of the square is
$$a = \frac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{2}} = 2 + \frac{\sqrt{2} + \sqrt{6}}{2}.$$