= Solution
Since $\mathbf B=0$,
$$
\mathcal E(r)=\frac{\epsilon_0E(r)^2}{2}
=\begin{cases}
\displaystyle\frac{q^2r^2}{32\pi^2\epsilon_0R^6},&r\leq R,\\[4pt]
\displaystyle\frac{q^2}{32\pi^2\epsilon_0r^4},&r>R.
\end{cases}
$$
Radial integration gives
$$
W_{\rm in}=\frac{q^2}{40\pi\epsilon_0R},
\qquad
W_{\rm out}=\frac{q^2}{8\pi\epsilon_0R},
$$
and therefore
$$
W_{\rm em}=\frac{3q^2}{20\pi\epsilon_0R}.
$$
To first order in $v/c$, $\mathbf E'=\mathbf E$ and $\mathbf B'=-\mathbf v\times\mathbf E/c^2$. Thus
$$
g'_z=\frac{\epsilon_0v}{c^2}E^2\sin^2\theta.
$$
Using the angular average $\langle\sin^2\theta\rangle=2/3$ and $\int\epsilon_0E^2dV=2W_{\rm em}$,
$$
P_z=\frac{4}{3}\frac{W_{\rm em}}{c^2}v
=\frac{q^2v}{5\pi\epsilon_0Rc^2}.
$$
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