= Solution
For an axial <spin> Hamiltonian $\mathcal H_D=D S_z^2$ with $D/k_B=-25$ K, the minima are $m_S=\pm5/2$. The highest levels along the reversal ladder are the $m_S=\pm1/2$ doublet. Hence
$$
\boxed{U/k_B=|D|\{(5/2)^2-(1/2)^2\}=25(6)=150\ \mathrm K,}
$$
about 1.25 kJ mol$^{-1}$ or 12.9 meV per molecule. Using $|D|S^2$ would incorrectly put a half-integer <spin> at a nonexistent $m_S=0$ level. Transverse anisotropy and <quantum tunnelling of magnetization> can create faster relaxation pathways, so this axial ladder barrier need not equal an experimentally fitted effective barrier.
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