= Solution
Apply the <curl of the curl identity> $\nabla\times(\nabla\times E)=\nabla(\nabla\cdot E)-\nabla^2E$ to <Faraday's law>, and use the <Ampère-Maxwell equation>. This gives
$$
\boxed{\frac{\partial^2E}{\partial t^2}-\nabla^2E=-\nabla\rho-\frac{\partial J}{\partial t}.}
$$
Applying the same identity to $B$, using $\nabla\cdot B=0$, gives
$$
\boxed{\frac{\partial^2B}{\partial t^2}-\nabla^2B=\nabla\times J.}
$$
For a spherically symmetric charge density, <Gauss's law> on a sphere of radius $r$ gives
$$
E(r)=\frac{\widehat{\mathbf r}}{r^2}\int_0^r\rho(s)s^2\,ds.
$$
For the unit-density ball of radius $a$,
$$
\boxed{
E=\begin{cases}\dfrac r3\widehat{\mathbf r},&r<a,\\[4pt]
\dfrac{a^3}{3r^2}\widehat{\mathbf r},&r\geq a,
\end{cases}
\qquad
\phi=\begin{cases}\dfrac{3a^2-r^2}{6},&r<a,\\[4pt]
\dfrac{a^3}{3r},&r\geq a.
\end{cases}}
$$
The potential is continuous at $r=a$ and tends to zero at infinity.
Let $\mathbf p=(1,0,0)$ and write $\mathbf1_{S_n}$ for the indicator of $S_n$. The layered density decomposes as
$$
\rho=\mathbf1_{S_0}+\sum_{n\geq1}2^{n-1}\mathbf1_{S_n}.
$$
The point $\mathbf p$ lies on every sphere $S_n$, whose radius is $R_n=4^{-n}$ and whose outward radial direction there is $\widehat{\mathbf x}$. A uniform ball of density $c$ contributes $cR_n/3$ to the field and $cR_n^2/3$ to the potential at its surface. Therefore
$$
E(\mathbf p)=\frac13\left(1+\sum_{n\geq1}2^{n-1}4^{-n}\right)\widehat{\mathbf x}
=\boxed{\frac12\widehat{\mathbf x}},
$$
and
$$
\phi(\mathbf p)=\frac13\left(1+\sum_{n\geq1}2^{n-1}16^{-n}\right)
=\boxed{\frac5{14}}.
$$
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