Intriguing Ornament, Part II

By symmetry, we may focus on one face of the cube. The surface area of a cap, $2\pi Rh=2\pi R(R-\frac{a}{2})$ , has to be equal to the area of a face of the cube, $a^2$ , minus the area of the disk covered up by the sphere, with radius $\sqrt{R^2-\frac{a^2}{4}}$ , by Pythagoras. Thus we need to solve the equation $2\pi R(R-\frac{a}{2})=a^2-\pi\left(R^2-\frac{a^2}{4}\right)$ . Division by $a^2$ produces the quadratic equation $2\pi q(q-\frac{1}{2})=1-\pi\left(q^2-\frac{1}{4}\right)$ , with the relevant solution $q=\frac{1}{6}+\sqrt{\frac{1}{9}+\frac{1}{3\pi}}\approx 0.6327$ . The answer is $\boxed{633}$ .