The electron configuration of F e 3 + is (atomic number of F e = 26 )

We know that atomic number of $\mathrm{Fe}$ is $26$ and its a $\mathrm{d}-$ block element and it belongs to $4\mathrm{th}$ period and $8\mathrm{th}$ group also the general electronic configuration for $\mathrm{d}-$block elements is $(\mathrm{n}-1){\mathrm{d}}^{1-10}{\mathrm{ns}}^{1-2}$ where $\mathrm{n}$ is the period number.

So, for the ferrous atom the configuration will be

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

Now, at the $3^{\mathrm{rd}}$ excited state the electronic configuration of ferrous ion will be

${\mathrm{Fe}}^{3+} = \left[\mathrm{Ar}\right]3{\mathrm{d}}^{5}4{\mathrm{s}}^0$