smc

Let the side length of the square be $s$. Let the top of the pyramid be $A$ and the square base be $BCDE$. Then, let the smaller cube meet edge $AB$ at $F$ and edge $AC$ at $G$. Let the closest vertice of the cube to $C$ on the base of the pyramid be $H$. Consider diagonal $CE$. It has length $\sqrt{2}$. Since the diagonal of the smaller cube's base is $s\sqrt{2}$, note that the distance from $C$ to $H$ is $\frac{\sqrt{2}-s\sqrt{2}}{2}.$ Also, note that $AG=s$ and $CG=1-s$. We can now use the Pythagorean Theorem on triangle $CHG$, the right angle at $G$, to solve for $s$. $$s^2+(\frac{\sqrt{2}-s\sqrt{2}}{2}{)}^2=(1-s{)}^2$$ $$\to s=\sqrt{2}-1\to \boxed{5\sqrt{2}-7}.$$