2005 AMC 10B Problem 7

Let the smaller circle have radius $r,$ so its area is $\pi r^2.$ The smaller square, which circumscribes this circle, has side $2r,$ and its diagonal $2\sqrt2\,r$ is the diameter of the larger circle. So the larger circle has radius $\sqrt2\,r.$ The larger square circumscribes the larger circle, so it has side $2\sqrt2\,r$ and area $8r^2.$ The desired ratio is $$\dfrac{\pi r^2}{8r^2} = \dfrac{\pi}{8}.$$ Thus, B is the correct answer.