jmc

Let $s$ be the side length of the square in Figure A.

Because the area of the semicircle in Figure A is half the area of the circle in Figure B, these two figures have the same radius, $r$. In figure A, if we draw a radius of the semicircle to a vertex of the inscribed square, we obtain a right triangle whose sides are $s/2$, $s$, and $r$. The Pythagorean Theorem tells us that $r^2=s^2+s^2/4$. After some manipulation, we see that $$s=\frac{2}{\sqrt{5}}r.$$ In Figure B, we see that the diameter of the circle makes up a diagonal of the square. Because the diagonal has length $2r$, it follows that the side length of the square is $2r/\sqrt{2}=r\sqrt{2}$.

To calculate the ratio of the areas, we square the ratio of the sides: $${\left(\frac{\frac{2r}{\sqrt{5}}}{r\sqrt{2}}\right)}^2={\left(\frac{2}{\sqrt{10}}\right)}^2=\frac{4}{10}=\boxed{\frac{2}{5}}.$$