= Solution
At weak field or high temperature, $\mu_BB\ll k_BT$ and $\tanh x\simeq x$, so
$$
M\simeq\frac{n\mu_B^2B}{k_BT}.
$$
Using $B\simeq\mu_0H$ for the applied-field response and neglecting demagnetizing corrections, the SI <volume magnetic susceptibility> is
$$
\boxed{\chi=\left.\frac{\partial M}{\partial H}\right|_{H=0}
=\frac{C}{T},\qquad C=\frac{\mu_0n\mu_B^2}{k_B}.}
$$
This is <Curie's law>. If susceptibility is instead defined as $\partial M/\partial B$, its Curie constant is $n\mu_B^2/k_B$ without $\mu_0$; stating which field is used prevents a units error.
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