= Solution
For this <molecular escape-speed temperature estimate>, $v_{\rm rms}^2=3k_BT/m$. Equating this to the surface escape-speed square $v_{\rm esc}^2=2GM_\oplus/R_\oplus$ gives
$$
\boxed{T=\frac{2mGM_\oplus}{3k_BR_\oplus}\simeq1.40\times10^5\,\mathrm K.}
$$
This evaluation keeps the paper's $R_\oplus=6.4\times10^6\,\mathrm m$ and $m=28u$, and uses externally documented $GM_\oplus=3.9860\times10^{14}\,\mathrm{m^3s^{-2}}$, $u=1.660539040\times10^{-27}\,\mathrm{kg}$ and $k_B=1.38064852\times10^{-23}\,\mathrm{J\,K^{-1}}$.<Earth Fact Sheet by National Aeronautics and Space Administration>[Bulk parameters: gravitational parameter GM]{ref}<CODATA recommended values of the fundamental physical constants 2014 by National Institute of Standards and Technology>[Atomic mass constant and Boltzmann constant]{ref} The corresponding escape speed is about $11.2\,\mathrm{km\,s^{-1}}$.
This is the requested speed-scale comparison. Atmospheric escape can occur from the high-speed tail at lower <temperature>; molecular dissociation and ionization also prevent treating an actual gas at this <temperature> as unchanged neutral nitrogen molecules.
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