= Solution
Let $g(E)$ denote the total <density of states>, including both <spin angular momentum>[spin] directions. The field lowers one spin energy by $\mu_BB$ and raises the other by the same amount. To first order in $B$, each spin population changes by the one-spin density $g(E_F)/2$ times the displacement of its Fermi edge. The population imbalance is therefore
$$
N_+-N_-
=2\mu_BB\frac{g(E_F)}2
=\mu_BB\,g(E_F).
$$
Multiplying by the <Bohr magneton> carried by each excess aligned electron gives
$$
\boxed{
M=\mu_B^2B\,g(E_F)
}.
$$
Thus, with this total-density convention,
$$
\boxed{\alpha=1,\qquad\gamma=2,\qquad\delta=1}.
$$
The zero-temperature <magnetic susceptibility> is
$$
\boxed{\chi=\frac{\partial M}{\partial B}
=\mu_B^2g(E_F)>0}.
$$
This weak positive response is <Pauli paramagnetism>. If susceptibility per unit volume is required, both $M$ and $g$ are divided by $V$.
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