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At zero temperature, fill all momentum states inside the Fermi sphere, with two spin states per momentum. Phase-space counting givesFor dispersion $E=cp$, the ratio of energy to particle-number integrals isLet $\epsilon$ be the common chemical potential in the weak field, and let $+$ denote the lower Zeeman-energy branch $cp-\mu B$. Its populations areKeeping $N_{+}+N_{-}=N$ changes $\epsilon$ only at order $B^2$, so to first order it equals ${\epsilon }_F=c{p}_F$. Subtracting the two populations givesEach excess lower-energy spin contributes moment $\mu$ along the field. Hence the Pauli moment of a massless electron gas isThis 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.