= Solution
Choose polar angle $\theta$ from the acceleration direction. Then $|n\times q\dot v|^2=q^2|\dot v|^2\sin^2\theta$. Integrating over the <solid angle> gives $2\pi\int_0^\pi\sin^3\theta\,d\theta=8\pi/3$, and hence the specified normalization gives <Larmor formula>
$$
\boxed{P=\frac{\mu_0q^2}{6\pi}|\dot v|^2.}
$$
This uses the source's convention for radiated power; standard SI normalization has an additional factor $1/c$ in both given power formulas.
To leading order when $W\ll E$, neglect the loss while computing the head-on trajectory. Mechanical energy conservation gives $\tfrac12m\dot r^2+V(r)=E=V(r_0)$ and acceleration magnitude $|V'(r)|/m$. On each leg,
$$
dt=\sqrt{\frac m2}\frac{|dr|}{\sqrt{V(r_0)-V(r)}}.
$$
The incoming and outgoing legs contribute equally. Multiplying the power by this time element, adding both legs, and assuming the integral converges gives
$$
\boxed{W\simeq\frac{\mu_0q^2}{3\pi m^2}\sqrt{\frac m2}
\int_{r_0}^\infty\frac{[V'(r)]^2}{\sqrt{V(r_0)-V(r)}}\,dr.}
$$
Using the unperturbed turning point is justified at this order in the small radiative loss, for a regular turning trajectory.
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