The spin-only magnetic moment of $\mathrm{Fe}^{3+}$ (ground state) is approximately:

$\mathrm{Fe}^{3+}: [\mathrm{Ar}]3d^5$, so $n = 5$ unpaired electrons (high-spin, weak-field assumption). $\mu = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92\,\text{BM}$. The other options correspond to $n = 1, 2$ and $4$ unpaired electrons.