= Solution
Pair equilibrium with negligible chemical potentials cannot continue once its predicted electron abundance falls below the residual abundance required by <charge neutrality>. Put $x=m_e/T$ and equate part (b), to logarithmic accuracy, to the order-$10^{-10}$ result of part (c):
$$
x^{3/2}e^{-x}\sim10^{-10}.
$$
Taking a logarithm and retaining the numerical factor singled out in the hint gives
$$
x-\frac32\log x\simeq10\log10\simeq23.
$$
The solution is $x\simeq28$, so
$$
\boxed{T\simeq\frac{m_e}{28}\simeq1.8\times10^{-2}\ {\rm MeV}\simeq20\ {\rm keV}}.
$$
Thus the zero-chemical-potential electron-positron equilibrium description must fail before roughly $20\,\mathrm{keV}$; thereafter the small asymmetric electron population survives.
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