= Solution
{c}
Treat the ordered spins as classical vectors of fixed length $S$, as in the requested mean-field energy. Each iron site has six nearest neighbors: four in the $xy$ plane and two along $z$. All are antiparallel in the collinear state. Each exchange bond contributes $-JS^2$ and is counted only once, giving $-(6/2)JS^2=-3JS^2$ per spin.
The easy-axis term contributes $-KS^2$ because $S_i^x=\pm S$. The cross product of two collinear spins is zero, so the term proportional to $D$ contributes nothing. Therefore
$$
\boxed{E_0=-S^2(3J+K).}
$$
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