= Solution
Periodic boundary conditions give wavevectors $(2\pi/L)(n_x,n_y)$. In the continuum, large-box limit each state occupies area $(2\pi/L)^2$ in wavevector space. A circular annulus therefore contains $[L^2/(2\pi)^2]2\pi k\,dk$ spinless states. Since $\epsilon=\hbar^2k^2/(2m)$, the <density of states> is
$$
\boxed{g_0=\frac{mL^2}{2\pi\hbar^2}.}
$$
At finite $L$ the exact spectrum is discrete; this formula is its continuum density.
With electron spin the two Zeeman branches are $\epsilon_{\mathrm{kin}}\mp\mu H$. The requested calculation includes this spin coupling and neglects orbital coupling to the field; for a perpendicular field, <Landau levels> would change the result. It is appropriate for a field parallel to the <ideal> two-dimensional plane, or for the stated spin-only model. Taking $\mu H\geq0$, the lower branch starts at $-\mu H$, and the upper one at $\mu H$, so
$$
\boxed{g(\epsilon)=\begin{cases}0,&\epsilon<-\mu H,\\g_0,&-\mu H<\epsilon<\mu H,\\2g_0,&\epsilon>\mu H.\end{cases}}
$$
Construct the ground state by filling the lowest one-electron levels subject to the <Pauli exclusion principle>. If the <Fermi energy> $\epsilon_F$ exceeds $\mu H$, the two populations are $N_+=g_0(\epsilon_F+\mu H)$ and $N_-=g_0(\epsilon_F-\mu H)$. Consequently
$$
\boxed{N=2g_0\epsilon_F=\frac{mL^2}{\pi\hbar^2}\epsilon_F,\qquad
M=\mu(N_+-N_-)=\frac{\mu^2mL^2}{\pi\hbar^2}H.}
$$
The <ground-state energy>, including the Zeeman term, follows by integrating the energy over each occupied branch:
$$
E_0=g_0\int_{-\mu H}^{\epsilon_F}\epsilon\,d\epsilon+
g_0\int_{\mu H}^{\epsilon_F}\epsilon\,d\epsilon
=g_0(\epsilon_F^2-\mu^2H^2)
=\boxed{\frac{\pi\hbar^2N^2}{2mL^2}-\frac12MH.}
$$
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