 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 $2\pi kdk$. ThereforeThe two-dimensional free-electron density of states is constant in energy, $g_2^{(1)}\propto {ϵ}^0$, above the band bottom. Including both spin states gives $g_2=L^{2}m/(\pi {\hslash }^2)$. Dividing by $L^2$ gives the density per unit physical area, rather than the total specimen's density.