= Solution
A larger <baryon-to-photon ratio> favours bound nuclei at a fixed <temperature> through the equilibrium factor $\eta^{A-1}$. In particular it helps deuterium survive photodissociation sooner: there are fewer energetic <photons> per <baryon>, and nuclear reaction rates per <baryon> also rise. A schematic <deuterium bottleneck> condition is $\eta e^{B_D/T_B}\sim\mathrm{constant}$, suppressing slowly varying powers of <temperature>. It gives
$$
\frac{B_D}{T_B}\simeq\mathrm{constant}-\ln\eta,\qquad
\frac{dT_B}{d\ln\eta}\simeq\frac{T_B^2}{B_D}>0.
$$
Thus higher $\eta$ permits burning at a higher <temperature>. During <radiation domination> that means earlier cosmic time, $t_B\propto T_B^{-2}$, leaving less time for <free-neutron decay after freeze-out>. More <neutrons> enter helium, so \b[the final helium-4 mass fraction increases], ordinarily rather weakly. This is the <baryon-density effect on primordial helium>.
The equilibrium helium factor $\eta^3$ indicates the direction of the change but is not a prediction that the final helium abundance grows without bound as $\eta^3$. Once almost all surviving <neutrons> are captured, the final <primordial helium mass fraction> is limited by their supply, as the shared <neutron> budget shows.
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