imc

Let the side of the largest square be $x$. It follows that the diameter of the inscribed circle is also $x$. Therefore, the diagonal of the square inscribed inscribed in the circle is $x$. The side length of the smaller square is $\frac{x}{\sqrt{2}}=\frac{x\sqrt{2}}{2}$. Similarly, the diameter of the smaller inscribed circle is $\frac{x\sqrt{2}}{2}$. Hence, its radius is $\frac{x\sqrt{2}}{4}$. The area of this circle is ${\left(\frac{x\sqrt{2}}{4}\right)}^2\pi =\frac{2\pi x^2}{16}=\frac{x^2\pi }{8}$, and the area of the largest square is $x^2$. The ratio of the areas is $\frac{\frac{x^2\pi }{8}}{x^2}=\frac{\cancel{x^2}\pi }{8}\cdot \frac{1}{\cancel{x^2}}=\boxed{\frac{\pi }{8}}$.