jmc

We label the points as shown. $A$ is the midpoint of the top side of the square, and $B$ is a vertex of the square. We look at right triangle $△OAB$. We seek a ratio of areas, which remains constant no matter the side lengths, so for simplicity, we let the big square have side length $2$ and the small square have side length $2x$. Then, $OA=1+2x$, $AB=x$, and $OB$ is a radius of the circle, which has length $\sqrt{2}$ by 45-45-90 triangles. Then, the Pythagorean theorem states that $O{A}^2+A{B}^2=O{B}^2$, or $$(1+2x{)}^2+x^2=(\sqrt{2}{)}^2.$$ Simplifying the equation yields $$\begin{array}{rl}& 1+4x+4x^2+x^2=2\\ ⟺& 5x^2+4x-1=0\\ ⟺& (5x-1)(x+1).\end{array}$$ Thus, $x=-1$ or $x=1/5$. Lengths are clearly positive, so the valid solution is $x=1/5$. Then the small square has side length $2x=2/5$, and area $(2/5{)}^2=4/25$. The large square has area $2^2=4$, so the small square has $$\frac{4/25}{4}=1/25=\boxed{4\%}$$ the area of the large square.