= Solution
Use the auxiliary <point charge> of strength $q\ne0$ at the origin, with potential $\phi_2=q/(4\pi\varepsilon_0r)$ from part (a). Apply the <electrostatic reciprocity identity> of part (b) to the exterior region $a<r<R$, then let $R\to\infty$. Its inner boundary has outward normal $-\mathbf e_r$, and $\rho_2=0$ everywhere in this exterior region. The outer boundary terms vanish under the stipulated decay assumption.
On $r=a$, $\phi_1=\Phi$ and
$$
\nabla\phi_2\cdot(-\mathbf e_r)=\frac{q}{4\pi\varepsilon_0a^2}.
$$
Thus the surviving boundary term on the left is $\Phi q/\varepsilon_0$. On the right, $\phi_2$ is constant on the inner sphere, so its boundary term is
$$
\frac{q}{4\pi\varepsilon_0a}\int_{r=a}\nabla\phi_1\cdot(-\mathbf e_r)\,dS=0.
$$
Indeed, $\nabla\phi_1=-\mathbf E_1$, and this integral is the outward electric flux from the interior ball, whose enclosed charge is zero by the hypothesis and <Gauss's law>. No symmetry of the exterior charge distribution is needed.
The <electrostatic reciprocity identity> now reduces to
$$
\frac{q\Phi}{\varepsilon_0}
=\frac1{\varepsilon_0}\int_{r>a}\frac{q\rho_1(\mathbf x)}{4\pi\varepsilon_0r}\,dV.
$$
Canceling $q/\varepsilon_0$ gives
$$
\boxed{\Phi=\frac1{4\pi\varepsilon_0}\int_{r>a}\frac{\rho_1(\mathbf x)}{r}\,dV.}
$$
The inward normal on the inner boundary is essential to the positive sign in this result.
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