Given the small inscribed square has an area of $8$, so a side length of $2\sqrt2$
The radius $r$ of the semicircle and circle is equal to the distance between the midpoint of the bottom side of the small inscribed square and one of the top vertices. This forms a right triangle with side lengths $2\sqrt2$, $\sqrt2$, and hypotenuse $r$. Using the Pythagorean theorem
$$(2\sqrt2)^2+(\sqrt2)^2=r^2$$
$$8+2=r^2$$
$$r=\sqrt{10}$$
Now that we have the radius of the circle, we know that the side length of the large inscribed square is $\frac{2r}{\sqrt2} = r\sqrt2$, ($2r$ is the diagonal of the large inscribed square, also the diameter of circle).
The side length of the large inscribed square is $\sqrt{20}=2\sqrt5$, so its area is $20$ square units.