= Solution
For a parabolic <two-dimensional electron gas> with one valley and two spin states, the <density of states> per area is $g_{2D}=m^*/(\pi\hbar^2)$. Integrating from the subband edge to the zero-temperature <Fermi energy> gives $n_s=g_{2D}E_F$, hence
$$
\boxed{E_F=\frac{\pi\hbar^2n_s}{m^*}=1.716\times10^{-21}\ \mathrm J
=10.71\ \mathrm{meV}.}
$$
Here $m^*=0.067(9.109\times10^{-31})=6.103\times10^{-32}\ \mathrm{kg}$, and the energy is measured from the occupied subband bottom. Equivalently $k_F=\sqrt{2\pi n_s}=1.3722\times10^8\ \mathrm{m^{-1}}$. Therefore
$$
\boxed{v_F=\frac{\hbar k_F}{m^*}=\sqrt{\frac{2E_F}{m^*}}
=2.371\times10^5\ \mathrm{m\,s^{-1}}.}
$$
The spin factor is already included in $g_{2D}$; counting it again would halve the energy incorrectly.
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