= Solution
Define $\mathcal H=-2J\sum_{\langle ij\rangle}\mathbf S_i\cdot\mathbf S_j$. Each site has $z=6$ exchange-connected neighbours. In <mean-field theory>, replacing neighbours by their mean <spin> produces the effective field that yields
$$
\Theta=\frac{2zJS(S+1)}{3k_B},\qquad\frac{J}{k_B}=\frac{\Theta}{4S(S+1)}.
$$
Thus
$$
\boxed{J/k_B\simeq(-0.414,\ -0.704,\ -3.55,\ -8.96)\ \mathrm K}
$$
for Mn, Fe, Co and Ni. Negative $J$ means antiferromagnetic exchange in this convention; a Hamiltonian written with $-J$ instead of $-2J$ would have twice these numerical values. The increasingly negative values indicate a growing effective exchange energy, although different orbital occupancy and anisotropy mean that they are not a simple universal function of $d$-electron number.
A three-dimensional six-connected framework is more favourable to mean-field reasoning than isolated chains, but the theory neglects short-range correlations, anisotropic exchange and fluctuations. The ratios $|\Theta|/T_C\simeq1.65,1.69,3.75,2.10$ show substantial departures from the simplest unfrustrated isotropic mean-field relation. The values are therefore \b[effective high-temperature exchange estimates], not a complete microscopic fit to the ordered phases.
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