= Solution
In two dimensions the <density of states> in $k$ space per <spin> species is $(L/2\pi)^2$. A <annulus> of radius $k$ and thickness $dk$ has area $2\pi k\,dk$. Therefore
$$
g_2^{(1)}(\epsilon)=\left(\frac{L}{2\pi}\right)^2 2\pi k\frac{dk}{d\epsilon}
=\boxed{\frac{L^2m}{2\pi\hbar^2}.}
$$
The <two-dimensional free-electron density of states> is constant in <energy>, $g_2^{(1)}\propto\epsilon^0$, above the band bottom. Including both <spin> states gives $g_2=L^2m/(\pi\hbar^2)$. Dividing by $L^2$ gives the density per unit physical area, rather than the total specimen's density.
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