I had an exam today and this old past paper question came up again that I had previously tried. I followed the hints and got an answer of $4π$, with the flux across v out of D being $4π$ and the triple integral over the divergence of v (by the divergence theorem) $8π$, hence the final answer being 8 π - 4 π = 4 π again by the divergence theorem. Is this the correct answer?
Thankyou
The flux across $D$ is $-4\pi$ (remember that the normal points downward when you think of $D$ as part of the boundary of $\Omega$). The triple integral of divergence over $\Omega$ is $3\cdot 16\pi/3 = 16\pi$. Thus, the flux across $S$ is $16\pi - (-4\pi) = 20\pi$. (Incidentally, this checks with a direct surface integral computation.)