Ground state of $Fe$: $[Ar]\ 3d^6\ 4s^2$. $Fe^{3+}$ loses 3 electrons (first from $4s$, then from $3d$): $[Ar]\ 3d^5$. In $3d^5$, all five $3d$ orbitals have one electron each (Hund's rule), giving $\mathbf{5}$ unpaired electrons.