Here's how I solved it:

Let the square be a unit square with vertices at (0, 0), (1, 0), (1, 1), (0, 1), and let the center of the small circle be (x, y) and the radius be r. Each of the 3 circular arcs can be used to form an equation relating to the distance from the centers of the circular arcs to the center of the small circle:

(0.5 - x)^2 + y^2 = (0.5 + r)^2

x^2 + (y - 0.5)^2 = (0.5 - r)^2

(1 - x)^2 + y^2 = (1 - r)^2

These equations can be rewritten as:

x^2 + y^2 - r^2 = r + x

x^2 + y^2 - r^2 = y - r

x^2 + y^2 - r^2 = 2x - 2r

Equating the linear expressions on the right sides of those equations gives x = 3r and y = 5r, and substituting this into the quadratic expression gives 33r^2 = 4r.