Notes
square in a circle in a quarter circle in a square solution

Square in a Circle in a Quarter Circle in a Square

Solution by Properties of Squares

In the above diagram, let $a$, $b$, and $c$ be the side lengths of the three squares in ascending order. As the smallest square has area $4$, its side length is $2$ so $a = 2$. The line segment $A B$ is a diagonal of the smallest square so its length is $2 \sqrt{2}$.

The diagonal of the outer square has length $c \sqrt{2}$. The radius of the quarter circle is $c$, so the length of $B E$ is $c$ and then of $A E$ is $c + 2 \sqrt{2}$. Putting these together shows that $c \sqrt{2} = c + 2 \sqrt{2}$ so $c = \frac{2 \sqrt{2}}{\sqrt{2} - 1}$.

The length of $O B$ is the same as that of $O D$, which is $\frac{b}{\sqrt{2}}$ as it is half a diagonal of the middle square. This then shows that $O E$ has length $b$ since it is the diagonal of a square with side length $\frac{b}{\sqrt{2}}$. So $B E$ has length $\frac{b}{\sqrt{2}} + b = b \frac{1 + \sqrt{2}}{\sqrt{2}}$. As this is the same as $c$, this gives the following expression for $b$:

Hence the area of the middle square is $16$.