= Solution
{c}
For the <thin spherical charge layer limit>, smoothness of $h$ gives
$$
\frac{h(1+\delta)-h(1)}\delta\longrightarrow h'(1).
$$
Thus the shell's total <electric charge> tends to $4\pi\rho_0a^3h'(1)$ while its thickness tends to zero. At each fixed $r>a$, the observation point eventually lies outside the shell, whereas at each fixed $r<a$ the interior <electric field> is unchanged. The limiting <electric field> away from the <sphere> is therefore
$$
\mathbf E_0(\mathbf r)=\begin{cases}
\dfrac{\rho_0}{3\epsilon_0}\mathbf r,&r<a,\\
\dfrac{\rho_0a^3}{\epsilon_0r^2}\left[\dfrac13+h'(1)\right]\mathbf e_r,&r>a.
\end{cases}
$$
Its outward-minus-inward jump is
$$
\boxed{\mathbf E_0(a^+)-\mathbf E_0(a^-)=\frac{\rho_0a h'(1)}{\epsilon_0}\mathbf e_r.}
$$
Physically, the limit is a uniformly charged <ball> of <charge density> $\rho_0$, together with an infinitesimally thin <spherical shell> of uniform <surface charge density>
$$
\boxed{\sigma=\rho_0a h'(1).}
$$
Indeed its <electric charge> is $4\pi a^2\sigma$, and the standard normal <electric field> jump is $\sigma/\epsilon_0$. The <surface charge density> can have either sign, or vanish, according to $h'(1)$.
\b[The order of limits matters.] For every finite $\delta>0$, taking the two one-sided limits at $r=a$ first gives zero jump. The displayed nonzero jump is obtained by first taking $\delta\to0$ at fixed points on either side, then approaching $a$. The value at the <surface> itself need not coincide with either trace of the limiting <electric field>.
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