= Solution
In the collisionless <exosphere>, an atom escapes if its outward trajectory has positive total mechanical <energy>. Neglect tides and stellar forces and use <Newtonian gravity> at exobase <radius> $r_e$:
$$
\frac12mv^2>\frac{GM_pm}{r_e},\qquad
v>v_{\rm esc}=\sqrt{\frac{2GM_p}{r_e}}.
$$
With thermal speed $v_{\rm th}=\sqrt{2k_BT_e/m}$, the <Jeans escape parameter> is
$$
\lambda_e=\frac{v_{\rm esc}^2}{v_{\rm th}^2}
=\frac{GM_pm}{k_BT_er_e}.
$$
A thermal distribution always has an escaping tail; <Jeans escape> is exponentially suppressed for $\lambda_e\gg1$, with <Jeans escape flux> proportional to $(1+\lambda_e)e^{-\lambda_e}$. Efficient escape requires $\lambda_e$ of order a few or smaller, with an order-unity energetic estimate $k_BT_e\sim GM_pm/r_e$.
Assume a <Neptune>-like <mass> and <radius>, $r_e\simeq R_p$, and atomic <hydrogen>. Using the supplied rounded constants gives
$$
g\simeq17.5\,\mathrm{m\,s^{-2}},\qquad
v_{\rm esc}\simeq2.65\times10^4\,\mathrm{m\,s^{-1}},\qquad
\frac{GM_pm_H}{k_BR_p}\simeq3.5\times10^4\,\mathrm K.
$$
Hence
$$
\boxed{T_e\sim10^4\text{--}4\times10^4\,\mathrm K
\text{ for a few-to-unity Jeans parameter at }r_e\simeq R_p.}
$$
Using the mean kinetic energy $3k_BT/2$ instead gives an order-unity coefficient and $T\simeq2.3\times10^4\,\mathrm K$. An expanded <exobase> has weaker binding and lowers the estimate by $R_p/r_e$. A comet-like tail can also be shaped by <radiation pressure> and stellar-wind interactions; it does not by itself measure $T_e$ or prove that a hydrostatic Jeans model is valid. At $\lambda_e\sim1$, $H/r_e\sim1$ and <hydrostatic equilibrium> fails as a global description: substantial mass loss must usually be treated as <hydrodynamic atmospheric escape>.
The <exobase> is defined by <mean free path> $\ell\sim H$, not by a universal <pressure>. For a neutral hydrostatic gas with <collision cross-section> $\sigma_c$,
$$
\ell\sim\frac1{n_e\sigma_c},\quad H\simeq\frac{k_BT_e}{m_Hg},\quad
\boxed{P_e=n_ek_BT_e\sim\frac{m_Hg}{\sigma_c}.}
$$
For example, explicitly assuming $\sigma_c\sim10^{-19}\text{--}10^{-20}\,\mathrm{m^2}$ gives $P_e\sim2\times10^{-7}\text{--}2\times10^{-6}\,\mathrm{Pa}$, or $2\times10^{-12}\text{--}2\times10^{-11}\,\mathrm{bar}$. These are representative extremely dilute neutral-exobase pressures, with orders of magnitude varying with composition, cross-sections and expansion. The supplied constants contain no collision information, so they cannot uniquely determine an exobase <pressure>; ionization or a non-hydrostatic density profile changes this estimate.
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