= Solution
<Chemical equilibrium> gives $\mu_{\rm He}=2\mu_p+2\mu_n$, since the photon <chemical potential> is zero. Set $A=(2\pi m_pkT/h^2)^{3/2}$. The degeneracies and approximate masses give prefactors $2A,2A,8A$ for protons, neutrons and helium. Thus
$$
\boxed{\frac{n_p^2n_n^2}{n_{\rm He}}=2A^3e^{-B_{\rm He}/kT}.}
$$
Here the binding energy uses $2m_p+2m_n-m_{\rm He}$; the PDF prints $2n_n$ in that mass relation, a dimensional typo.
Baryon conservation gives $n_B\propto a^{-3}$. In adiabatic expansion without photon-producing entropy release, $T\propto a^{-1}$ and $n_\gamma\propto T^3\propto a^{-3}$, so \b[$\eta=n_B/n_\gamma$ stays constant]. This statement is for that epoch; entropy transfer such as electron-positron annihilation modifies the photon normalization.
Since the density ratio equals $n_B^3X_p^2X_n^2/X_{\rm He}$, substitution of the given photon density yields
$$
\boxed{\frac{X_p^2X_n^2}{X_{\rm He}}
=\frac{\eta^{-3}}{32}\left(\frac\pi2\right)^{3/2}
\left(\frac{m_pc^2}{kT}\right)^{9/2}e^{-B_{\rm He}/kT}.}
$$
At $m_pc^2\simeq938$ MeV, $B_{\rm He}=30$ MeV and $kT=0.3$ MeV, the right side is about $1.22\times10^{-29}\eta^{-3}$. Unity therefore corresponds to $\eta\simeq2.3\times10^{-10}$, illustrating the stated very rough temperature estimate for a cosmological <baryon-to-photon ratio> of order $10^{-10}$ to $10^{-9}$. A numerical value of $\eta$ is necessary; it is not specified in the question. The exponential makes the temperature estimate far less sensitive than the ratio itself to order-one prefactors.
Actual helium synthesis must proceed through light intermediates. The <deuterium bottleneck> persists because deuterium's much smaller binding energy allows the abundant high-energy photon tail to destroy it. Only at a lower temperature do enough deuterons survive to initiate efficient reactions leading to helium. Direct four-nucleon assembly at the <equilibrium> temperature is ineffective.
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