jmc

Let the radius of the smaller circle be $r$. Then the side length of the smaller square is $2r$. The radius of the larger circle is half the length of the diagonal of the smaller square, so it is $\sqrt{2}r$. Hence the larger square has sides of length $2\sqrt{2}r$. The ratio of the area of the smaller circle to the area of the larger square is therefore $$\frac{\pi r^2}{{\left(2\sqrt{2}r\right)}^2}=\boxed{\frac{\pi }{8}}.$$