$MnO _{4}^{-}+5 e^{-} \rightarrow Mn ^{2+}$

The reaction equation of compound $FeC _{2} O _{4}$ is,

$FeC _{2} O _{4} \rightarrow Fe ^{3+}+2 CO _{2}+3 e^{-}$

So, $1$ mole of $FeC _{2} O _{4}$ requires $\frac{3}{5}$ moles of $KMnO _{4}$.

The reaction equation of compound $Fe _{2}\left( C _{2} O _{4}\right),$ is,

$Fe _{2}\left( C _{2} O _{4}\right)_{3} \rightarrow Fe ^{3+}+ CO _{2}+6 e^{-}$

So, $1$ mole of $Fe _{2}\left( C _{2} O _{4}\right)_{3}$ requires $\frac{6}{5}$ moles of $KMnO _{4}$.

The reaction equation of compound $FeSO _{4}$ is,

$FeSO _{4} \rightarrow Fe ^{3+}+e^{-}$

So, $1$ mole of $FeSO _{4}$ requires $\frac{1}{5}$ moles of $KMnO _{4}$.

since $Fe _{2}\left( SO _{4}\right)_{3}$ does not oxidize.

The total number of required moles is,

$n=\frac{3}{5}+\frac{6}{5}+\frac{1}{5}$

$n=2$