 Solution

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 $-J{S}^2$ and is counted only once, giving $-(6/2)J{S}^2=-3J{S}^2$ per spin.

The easy-axis term contributes $-K{S}^2$ because $S_i^x=\pm S$. The cross product of two collinear spins is zero, so the term proportional to $D$ contributes nothing. Therefore