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 $A B C$ - hence, the desired radius is merely the circumradius of triangle $A B C$ minus $\frac{1}{2}$. Now note that by the Pythagorean theorem, $B C=\sqrt{2}$ and $A B=A C=\sqrt{5}$ so we easily find that the altitude from $A$ in triangle $A B C$ has length $\frac{3 \sqrt{2}}{2}$. Therefore the area of triangle $A B C$ is $\frac{3}{2}$. Hence the circumradius of triangle $A B C$ is given by $$\frac{B C \cdot C A \cdot A B}{4 \cdot \frac{3}{2}}=\frac{5 \sqrt{2}}{6}$$ and so the answer is $\frac{5 \sqrt{2}}{6}-\frac{1}{2}=\frac{5 \sqrt{2}-3}{6}$.