Which is the electronic configuration for an $Fe\left(IIl\right)$ ion in its ground state?

The electronic configuration for an $Fe\left(IIl\right)$ ion in its ground state is $\left[\mathrm{Ar}\right]3d^5$

Arrangement of Fe in electronic terms,

$\mathrm{Fe}=\left[\mathrm{Ar}\right]3{\mathrm{d}}^{6}4{\mathrm{s}}^2$

In this case, Fe's electronic configuration (III),

${\mathrm{Fe}}^{+3}=\left[\mathrm{Ar}\right]3{\mathrm{d}}^5$

Hence, option A is correct.