= Solution
{c}
Set $r=R$ in the enclosed <electric charge> formula:
$$
\boxed{Q=4\pi\rho_0a^3\left[\frac13+\frac{h(1+\delta)-h(1)}{\delta}\right].}
$$
Outside the entire <charge distribution>, <Gauss's law> gives $E_r(r)=Q/(4\pi\epsilon_0r^2)$. Integrating $E_r=-\varphi'$ with $\varphi(\infty)=0$ gives
$$
\boxed{\varphi(r)=\int_r^\infty\frac{Q}{4\pi\epsilon_0s^2}\,ds
=\frac{Q}{4\pi\epsilon_0r},\qquad r>R.}
$$
Thus the exterior <electric potential> and <electric field> are exactly those of a <point charge> $Q$ placed at the center. This is the <electrostatic shell theorem>: outside a <spherically symmetric charge distribution>, only its total <electric charge> affects the <electric field>, not how that <electric charge> is distributed radially. If positive and negative <electric charges> cancel to give $Q=0$, both exterior quantities vanish.
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