= Solution
For one spin-degenerate parabolic two-dimensional band, $k_F=\sqrt{2\pi n}=1.3729\times10^8$ m$^{-1}$. The <Free electron model>[free-electron model] with $m=m_e$ gives
$$
\boxed{E_F=\frac{\pi\hbar^2n}{m_e}=7.182\times10^{-4}\ \mathrm{eV}
=0.7182\ \mathrm{meV},\qquad T_F=8.334\ \mathrm K.}
$$
The given $10^3\ \Omega$ is a <sheet resistivity>, not a bulk <Electrical resistivity>[resistivity in ohm-metres]. The two-dimensional <Drude model>[Drude] <Electrical conductance>[sheet conductance] is $ne^2\tau/m_e$, so
$$
\boxed{\tau=\frac{m_e}{ne^2\rho_{\square}}=1.183\times10^{-11}\ \mathrm s.}
$$
Thus $v_F=\hbar k_F/m_e=1.589\times10^4$ m/s and
$$
\boxed{\ell_F=v_F\tau=1.880\times10^{-7}\ \mathrm m=188\ \mathrm{nm}.}
$$
The sample <temperature> is not given. This is specifically the \b[Fermi-surface] <mean free path>, appropriate in the degenerate regime $T\ll8.3$ K for the bare-mass model. At higher <Temperature>[temperatures] the characteristic <Thermal speed>[thermal speed] should replace $v_F$ for a thermal transport distance. If a real two-dimensional band has mass $m^*$, the stated results scale as $E_F,T_F\propto1/m^*$, $\tau\propto m^*$ and $v_F\propto1/m^*$; the inferred $\ell_F=\hbar k_F/(ne^2\rho_{\square})$ is mass independent under these same <Drude model>[Drude] assumptions. No semiconductor <effective mass> or valley degeneracy is silently introduced.
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