 Solution

Use radiation-era $H=C_{\ast }{T}^2/m_{\mathrm{pl}}$, with $C_{\ast }=1.66\sqrt{g_{\ast }}$. The stipulated interaction rate is ${\alpha }_Y^4{T}^5/m_Y^4$, not ${\alpha }_Y^2{T}^5/m_Y^4$. Decoupling therefore givesAt the threshold $T_D=m_{Y\mathrm{nr}}$, the assumed $g_{\ast }\simeq 100$ is self-consistent with the stated high-temperature condition. Smaller masses decouple nonrelativistically, with $x_D=m_Y/T_D=(m_{Y\mathrm{nr}}/m_Y{)}^{1/3}\gg 1$.

For an thermal equilibrium species with zero chemical potential, $n_Y=2(m_Y{T}_D/2\pi {)}^{3/2}{e}^{-x_D}$, while $s=2{\pi }^2{g}_{\ast s}{T}_D^3/45$. ThusThe numerical coefficient before rounding is about $2.9\times 10^{-3}$ for $g_Y=2$ and $g_{\ast s}=100$. This is mass-dependent fermion freeze-out. It assumes the rate maintains chemical equilibrium until sudden cosmological particle freeze-out and that entropy is conserved afterwards. Elastic scattering alone would not justify the thermal equilibrium number-density estimate. For decoupling below the high-temperature range, changes in effective degrees of freedom modify the numerical estimate; the requested formula treats them as roughly constant.