= Solution
Fix the <electric potential> to vanish at infinity and assume a localized, nonsingular <charge density> of finite <electrostatic energy>. Assemble the charge by scaling $\rho_s=s\rho$ from $s=0$ to $1$. Linearity gives $\phi_s=s\phi$, so the work is
$$
U=\int_0^1ds\int\phi_s\rho\,d^3x
=\frac12\int\rho\phi\,d^3x.
$$
The factor $1/2$ prevents double counting of pair interactions. Using <Gauss law>, $\rho=-\epsilon_0\Delta\phi$, and <integration by parts>,
$$
U=-\frac{\epsilon_0}2\int\phi\Delta\phi\,d^3x
=\boxed{\frac{\epsilon_0}2\int|\nabla\phi|^2\,d^3x
=\frac{\epsilon_0}2\int|\mathbf E|^2\,d^3x}.
$$
The boundary integral at infinity vanishes for a localized distribution since $\phi=O(r^{-1})$ and $\partial_r\phi=O(r^{-2})$. Singular point-charge self-energies require a separate regularization and are not finite instances of this identity.
For the uniform solid sphere, spherical symmetry and <Gauss law> give
$$
\boxed{\mathbf E(r)=
\begin{cases}
\dfrac{\rho r}{3\epsilon_0}\,\widehat{\mathbf r},&0\le r\le R,\\[2pt]
\dfrac{\rho R^3}{3\epsilon_0r^2}\,\widehat{\mathbf r},&r\ge R.
\end{cases}}
$$
Integrating inward from infinity and matching the continuous <electric potential> at $R$ gives
$$
\boxed{\phi(r)=
\begin{cases}
\dfrac{\rho}{6\epsilon_0}(3R^2-r^2),&r\le R,\\[2pt]
\dfrac{\rho R^3}{3\epsilon_0r},&r\ge R.
\end{cases}}
$$
The value of the <electric field> at the centre is the zero vector.
Writing $Q=4\pi\rho R^3/3$, the <electrostatic energy of a uniformly charged solid sphere> follows either from the charge-potential integral or from the field integral including the exterior:
$$
U=\frac12\,4\pi\rho\int_0^R\phi(r)r^2\,dr
=\boxed{\frac{4\pi\rho^2R^5}{15\epsilon_0}
=\frac{3Q^2}{20\pi\epsilon_0R}}.
$$
For the nuclear scaling, $Q=Ze$ while constant volume per proton gives $R\propto Z^{1/3}$. Therefore \b[the electric contribution scales as $Q^2/R\propto Z^{5/3}$] in this uniform-density model.
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