Let $x$ be the side length of the square tile. If the rectangle is completely tiled with such square tiles, there exist integers $a, b$ such that the side lengths of the rectangle are $ax$ and $bx$. We let $r = 75$ denote the radius of the circle.

The diagonals of the rectangle are diameters of the circle and so their length is $2r$. Applying Pythagoras to the right angled triangle obtained by cutting the rectangle along one of its diagonals, we obtain
$$(2r)^2 = x^2(a^2 + b^2).$$
As $x, r, a, b$ are integers, this implies that $x$ is a factor of $2r$ and so $c = 2r/x$ is an integer that satisfies $c^2 = a^2 + b^2$. Because we are to find the largest possible $x = 2r/c$, we are interested in the smallest $c$ that divides $2r$ and which appears as the hypotenuse of a right angled triangle with integer side lengths.

The positive divisors of $2r = 150 = 2 \cdot 3 \cdot 5^2$ are $1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75$ and $150$. Because $1^2 + 1^2 = 2$, $1^2 + 2^2 = 5$, $2^2 + 2^2 = 8$ and all other sums of two positive squares are greater than $9$, we confirm that none of the numbers $1^2 = 1$, $2^2 = 4$ and $3^2 = 9$ is the sum of two non-zero squares. This rules out $1, 2, 3$ for $c$. Therefore, $c$ is at least $5$.

When $c = 5$, we can use $a = 3$ and $b = 4$ to get $c^2 = a^2 + b^2$. In this case we obtain $x = 150/5 = 30$. The side lengths of the rectangle are then equal to $ax = 90$ and $bx = 120$. Such a rectangle is indeed inscribed into a circle of radius $r = 75$, because $150^2 = 90^2 + 120^2$ which is just $30^2$ times $5^2 = 3^2 + 4^2$.

Therefore, the largest possible integer side length of a square tile that can be used to completely tile a rectangle that is inscribed in a circle of radius $75$ is equal to $30$.