Solution:

Let $A$ be the center of the square in the lower left corner, let $B$ be the center of the square in the middle of the top row, and let $C$ be the center of the rightmost square in the middle row. It's clear that $O$ is the circumcenter of triangle $ABC$—hence, the desired radius is merely the circumradius of triangle $ABC$ minus $\frac{1}{2}$. Now note that by the Pythagorean theorem, $BC = \sqrt{2}$ and $AB = AC = \sqrt{5}$ so we easily find that the altitude from $A$ in triangle $ABC$ has length $\frac{3\sqrt{2}}{2}$. Therefore the area of triangle $ABC$ is $\frac{3}{2}$. Hence the circumradius of triangle $ABC$ is given by
$$\frac{BC \cdot CA \cdot AB}{4 \cdot \frac{3}{2}} = \frac{5\sqrt{2}}{6}$$
and so the answer is $\frac{5\sqrt{2}}{6} - \frac{1}{2} = \frac{5\sqrt{2} - 3}{6}$.