 Solution

Let $a=h/2$ and place the hole axis along $z$. At the rim, $r_0^2=R^2-a^2$. A horizontal slice at $|z|\le a$ is an annulus of area $\pi [(R^2-z^2)-r_0^2]=\pi (a^2-z^2)$. Thus the napkin ring problem givesEach slice depends only on $a$ and $z$, explaining the independence from the original sphere radius, rather than obtaining it merely by cancellation after integration.