= Solution
Specify the <spin> counting first. Let $g_s=g_s(E_F)$ denote the unpolarized <single-spin density of states> per atom per <energy>, and put $\delta n=n_\uparrow-n_\downarrow$. At fixed total occupation $n$, the two <Hubbard model> mean-field energies can be written
$$
\varepsilon_{\mathbf k,\uparrow}=\varepsilon_{\mathbf k}+\frac{Un}{2}-\Delta,\qquad \varepsilon_{\mathbf k,\downarrow}=\varepsilon_{\mathbf k}+\frac{Un}{2}+\Delta,\qquad \Delta=\frac{U\delta n}{2}+\mu_0\mu_BH.
$$
The common shift $Un/2$ can be absorbed into the <chemical potential>. Lowering the up-spin band by $\Delta$ adds $g_s\Delta$ carriers, while raising the down-spin band removes the same number. Thus
$$
\delta n=2g_s\Delta=Ug_s\delta n+2\mu_0\mu_Bg_sH,\qquad \delta n=\frac{2\mu_0\mu_Bg_s}{1-Ug_s}H.
$$
If $n_a$ is the atom <number density>, the <magnetization> is $M=n_a\mu_B\delta n$. The <Stoner-enhanced magnetic susceptibility> in SI volume units is therefore
$$
\boxed{\chi=\frac{\partial M}{\partial H}=\frac{2\mu_0n_a\mu_B^2g_s(E_F)}{1-Ug_s(E_F)}=\frac{\chi_P}{1-Ug_s(E_F)}.}
$$
Here $\chi_P$ is the noninteracting <Pauli paramagnetism> result. If the paper's $g(E_F)$ counts both spins, $g=2g_s$ and the same result is
$$
\boxed{\chi=\frac{\mu_0n_a\mu_B^2g(E_F)}{1-Ug(E_F)/2}.}
$$
A susceptibility per atom omits $n_a$. Stating the <spin> and volume conventions avoids an otherwise spurious factor of two or a dimensional mismatch.
The <Stoner criterion> is $Ug_s(E_F)>1$, equivalently $Ug(E_F)>2$ with spin-summed $g$. To see the instability, transfer $\eta$ <Electrons> per atom from one <spin> band to the other. The band-filling cost is $\eta^2/g_s$, whereas the interaction <energy> changes by $-U\eta^2$. Their sum $(1/g_s-U)\eta^2$ becomes negative at the threshold. The unpolarized state then ceases to be a local minimum, and nonlinear band filling determines the spontaneous <ferromagnetism>. The negative continuation of the displayed linear susceptibility beyond the threshold is not a stable paramagnetic response.
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