= Solution
{c}
Take $v=|\mathbf x|^2$ in <Green's second identity>. Since $\nabla^2u=0$, $\nabla^2v=6$, $v=1$ on the unit <sphere>, and $\partial_nv=2$ there,
$$
6\iiint_Bu\,dV=2\iint_{\partial B}u\,dS-\iint_{\partial B}\partial_nu\,dS.
$$
The final term vanishes by the <divergence theorem>, because it equals $\iiint_B\nabla^2u\,dV=0$. Therefore
$$
\boxed{\iiint_Bu\,dV=\frac13\iint_{\partial B}u\,dS.}
$$
We assume sufficient boundary regularity for the stated boundary values and <integrals>, or use the corresponding trace/<limit> formulation. This is the equality of <volume> and surface averages for a <harmonic function>, since the <unit ball> and <sphere> have <volume> $4\pi/3$ and <area> $4\pi$.
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