= Solution
At zero temperature, fill all momentum states inside the Fermi sphere, with two <spin> states per momentum. Phase-space counting gives
$$
N=2\frac V{(2\pi\hbar)^3}\frac{4\pi p_F^3}{3},\qquad
\boxed{p_F=\hbar(3\pi^2N/V)^{1/3}.}
$$
For dispersion $E=cp$, the ratio of energy to particle-number integrals is
$$
\frac{E_{tot}}N=c\frac{\int_0^{p_F}p^3dp}{\int_0^{p_F}p^2dp}
=\boxed{\frac34cp_F.}
$$
Let $\varepsilon$ be the common <chemical potential> in the weak field, and let $+$ denote the lower Zeeman-energy branch $cp-\mu B$. Its populations are
$$
N_\pm=\frac V{6\pi^2\hbar^3c^3}(\varepsilon\pm\mu B)^3.
$$
Keeping $N_++N_-=N$ changes $\varepsilon$ only at order $B^2$, so to first order it equals $\varepsilon_F=cp_F$. Subtracting the two populations gives
$$
N_+-N_-=\frac{3N\mu B}{\varepsilon_F}+O(B^3).
$$
Each excess lower-energy <spin> contributes moment $\mu$ along the field. Hence the <Pauli moment of a massless electron gas> is
$$
\boxed{\mathcal M=\frac{3N\mu^2B}{cp_F}+O(B^3).}
$$
This includes the <spin> contribution specified in the problem. The field is weak enough that both <spin> branches remain occupied, and orbital magnetic effects are not part of the assumed energy model.
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