= Solution
To preserve the convention of part (a), first count one <spin> species. Define $g_d^{(1)}(\epsilon)\,d\epsilon$ as the number of spatial states in the whole specimen in that <energy> interval. Put
$$
k=\frac{\sqrt{2m\epsilon}}{\hbar},\qquad
\frac{dk}{d\epsilon}=\frac{m}{\hbar^2k}.
$$
In one dimension there are intervals near both $+k$ and $-k$, so the <one-dimensional free-electron density of states> is
$$
g_1^{(1)}(\epsilon)=\frac{L}{2\pi}\,2\frac{dk}{d\epsilon}
=\boxed{\frac{L\sqrt{2m}}{2\pi\hbar}\epsilon^{-1/2}.}
$$
Thus $g_1^{(1)}\propto\epsilon^{-1/2}$. The factor of two here comes from opposite <momenta>, not <spin>. For physical <Electrons> with two <spin> states, the <spin-summed density of states> is $g_1=2g_1^{(1)}=L\sqrt{2m}\,\epsilon^{-1/2}/(\pi\hbar)$. These expressions are continuum approximations away from individual discrete levels.
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