= Solution
The specified two-direction constraint permits magnetic moments $\mu_z=\pm\mu$, with
$$
\mu=g_JS\mu_B=7\mu_B,
$$
where $\mu_B$ is the <Bohr magneton>. There are four magnetic <gadolinium> ions per conventional <face-centered cubic lattice> cell; the closed-shell nitrogen ions add no local spin moment. Hence the magnetic-ion <number density> is
$$
n=\frac4{a^3}=3.20\times10^{28}\,\mathrm{m^{-3}}.
$$
At high <temperature>, neglect interactions and use $B_z\simeq\mu_0H_z$. The one-ion <partition function> and mean <magnetic dipole moment> are
$$
Z_1=2\cosh\!\left(\frac{\mu\mu_0H_z}{k_BT}\right),\qquad
\langle\mu_z\rangle=\mu\tanh\!\left(\frac{\mu\mu_0H_z}{k_BT}\right).
$$
In the weak-<magnetic field> limit, $\tanh x\simeq x$, so $M_z=n\langle\mu_z\rangle=\mu_0n\mu^2H_z/(k_BT)$. Thus the dimensionless SI <magnetic susceptibility> is $\chi=C/T$, with
$$
\boxed{C=\frac{\mu_0n(7\mu_B)^2}{k_B}=12.27\,\mathrm K.}
$$
The coefficient has units of kelvin. The unrestricted-spin expression proportional to $S(S+1)/3$ is inappropriate here: the source explicitly retains only the two orientations with $S_z=\pm S$. In this <two-state Curie law>, the squared axial moment is $(g_JS\mu_B)^2$.
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