Solution
= Solution
Set $F=\omega_0e^{-\Delta E/(k_BT)}$ and $K=nQ^2F/(2k_BT)$. The six simple cubic neighbour vectors are $\pm a\hat x$, $\pm a\hat y$, $\pm a\hat z$. Each diagonal direction receives $2a^2$, and every off-diagonal product vanishes. Therefore
$$
\boxed{\boldsymbol\sigma_{\rm sc}=2Ka^2 I=\frac{nQ^2Fa^2}{k_BT}I}.
$$
The result uses $a$ as the nearest-neighbour distance specified here, not automatically as the conventional-cell edge for the other lattices.