a
Consider the reaction,

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

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,

\(n=\frac{3}{5}+\frac{6}{5}+\frac{1}{5}\)

\(n=2\)