= Solution
Fully ionized <helium> supplies three particles per four <hydrogen> <masses>, so $\mu_{\rm He}\simeq4/3$. The same small $Z$, but negligible <hydrogen> fraction, gives $\kappa_{0,\rm He}\propto Z$, half the hydrogen-star coefficient. The relative <luminosity> is
$$
\frac{L_{\rm He}}{L_H}
=\frac{\kappa_{0,H}}{\kappa_{0,\rm He}}
\left(\frac{4/3}{1/2}\right)^7
\left(\frac{0.5}{1}\right)^5
\left(\frac{10^8}{1.5\times10^7}\right)^{1/2}.
$$
Consequently
$$
\boxed{L_{\rm He}/L_H\simeq2\left(\frac83\right)^7\frac1{32}\sqrt{\frac{20}3}
\simeq155.}
$$
The very strong molecular-weight dependence outweighs the reduction in <mass>. Treating the two <opacities> as identical would miss a factor two, and using neutral-gas molecular weights would be inconsistent with the stated hot ionized <stars>.
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