= Solution
Using the supplied <iron> parameters gives
$$
C=\frac{\mu_0(8.5\times10^{28})(2\mu_B)^2(1)(2)}{3k_B}\simeq1.77\,\mathrm K,\qquad
\boxed{\lambda=\frac{1043}{C}\simeq5.88\times10^2.}
$$
An internal <Weiss molecular field> depends on <temperature> through $M(T)$. Its low-temperature saturation estimate uses $M_s=ng\mu_BJ\simeq1.58\times10^6\,\mathrm{A\,m^{-1}}$:
$$
\boxed{B_{e,\rm sat}=\mu_0\lambda M_s=\frac{3k_BT_C}{g\mu_B(J+1)}\simeq1.16\times10^3\,\mathrm T.}
$$
At and above $T_C$ the zero-applied-field <magnetization> vanishes in this model, so this is a saturation estimate, not the field at $T_C$.
Use the paper's stipulated <face-centered cubic lattice>, which has twelve nearest neighbours. The saturated one-ion alignment scale is
$$
(g\mu_BJ)B_{e,\rm sat}=\frac{3J}{J+1}k_BT_C\simeq0.135\,\mathrm{eV}.
$$
Dividing the effective field among twelve equivalent neighbours gives the per-bond <exchange interaction> scale
$$
\boxed{E_{\rm bond}\simeq\frac{g\mu_BJ B_{e,\rm sat}}{12}\simeq1.12\times10^{-2}\,\mathrm{eV}=11.2\,\mathrm{meV}.}
$$
To specify the convention, write the <Heisenberg model> as $\mathcal H=-\mathcal J\sum_{\langle ij\rangle}\mathbf J_i\cdot\mathbf J_j$, with dimensionless angular-momentum operators and each bond counted once. Then $g\mu_BB_e=12\mathcal J\langle J_z\rangle$, so $\mathcal J=3k_BT_C/[12J(J+1)]\simeq11.2\,\mathrm{meV}$ here. A convention with $-2\mathcal J'$ instead has $\mathcal J'=5.62\,\mathrm{meV}$. The total saturated interaction energy per ion is $-12\mathcal J J^2/2\simeq-67.4\,\mathrm{meV}$; the factor of one half avoids counting each bond twice.
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