= Solution
{c}
By the assumed <spherical symmetry> of the <electric potential>, the <electric field> is radial: $\mathbf E=E_r(r)\mathbf e_r$ for $r>0$. On a <sphere> of radius $r$, <Gauss's law> therefore reads
$$
4\pi r^2E_r(r)=\frac{Q(r)}{\epsilon_0},\qquad
Q(r)=4\pi\int_0^r\rho(s)s^2\,ds.
$$
Put $R=a(1+\delta)$. For $r<a$, the enclosed <electric charge> is $4\pi\rho_0r^3/3$. In the <spherical shell>, the factor $s^2$ in the <volume element> cancels $s^{-2}$ in the <charge density>, so
$$
\begin{aligned}
Q(r)&=\frac{4\pi\rho_0a^3}{3}
+\frac{4\pi\rho_0a^2}{\delta}\int_a^rh'(s/a)\,ds\\
&=4\pi\rho_0a^3\left[\frac13+\frac{h(r/a)-h(1)}{\delta}\right].
\end{aligned}
$$
It follows that
$$
\boxed{\mathbf E(\mathbf r)=\begin{cases}
\dfrac{\rho_0}{3\epsilon_0}\mathbf r,&0\leq r<a,\\
\dfrac{\rho_0a^3}{\epsilon_0r^2}\left[\dfrac13+\dfrac{h(r/a)-h(1)}{\delta}\right]\mathbf e_r,&a\leq r<R,\\
\dfrac{\rho_0a^3}{\epsilon_0r^2}\left[\dfrac13+\dfrac{h(1+\delta)-h(1)}{\delta}\right]\mathbf e_r,&r\geq R.
\end{cases}}
$$
At the origin the first formula gives $\mathbf E=\mathbf0$ without needing a radial <unit vector>. At both shell interfaces the adjacent formulas agree: for every fixed $\delta>0$, there is no <surface charge> concentrated on either <sphere>, and the <electric field> is continuous there. No sign restriction on $h'$ has been assumed, so the shell <charge density> and the radial <electric field> can be negative.
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