= Solution
Assemble the <spherically symmetric charge distribution> by adding thin concentric shells, with the final <charge density> kept uniform. When the assembled radius is $s$, the enclosed <electric charge> and the next increment are
$$
q(s)=Q\frac{s^3}{r^3},\qquad dq=\frac{3Qs^2}{r^3}\,ds.
$$
The <electric potential> of the already assembled material at that shell is $k_eq(s)/s$. Consequently its assembly work is $dU=k_eq(s)\,dq/s$, and the <electrostatic energy of a uniformly charged solid sphere> is
$$
U=\frac{3k_eQ^2}{r^6}\int_0^r s^4\,ds,
\qquad
\boxed{U=\frac{3k_eQ^2}{5r}=\frac{3Q^2}{20\pi\epsilon_0r}.}
$$
There is no extra factor of one half in this assembly <integral>: each pair of charge elements has already been counted exactly once, when the later element was brought in. A solid uniformly charged insulator has this value; a conducting sphere has its <electric charge> on the surface and instead has <electrostatic energy> $k_eQ^2/(2r)$.
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