Solution:
Let $r$ be the radius of the sphere. We take cross sections of the bead perpendicular to the line of the drill and compare them to cross sections of the sphere at the same distance from its center. At a height $h$, the cross section of the sphere is a circle with radius $\sqrt{r^{2}-h^{2}}$ and thus area $\pi\left(r^{2}-h^{2}\right)$. At the same height, the cross section of the bead is an annulus with outer radius $\sqrt{13^{2}-h^{2}}$ and inner radius $5$, for an area of $\pi\left(13^{2}-h^{2}\right)-\pi\left(5^{2}\right)=\pi\left(12^{2}-h^{2}\right)$ (since $13^{2}-5^{2}=12^{2}$). Thus, if $r=12$, the sphere and the bead will have the same cross-sectional area $\pi\left(12^{2}-h^{2}\right)$ for $|h| \leq 12$ and $0$ for $|h|>12$. Since all the cross sections have the same area, the two clay figures then have the same volume. And certainly there is only one value of $r$ for which the two volumes are equal, so $r=12$ is the answer.