$\mathrm{KMaO}_{4}$ is $5$

(a) For $\mathrm{FeSO}_{3}$ $\left.\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+} \quad \text { (No. of } e^{-} \mathrm{s} \text { involved }=1\right)$

$\mathrm{SO}_{3}^{2-} \longrightarrow \mathrm{SO}_{4}^{2-} \quad\left(\mathrm{No} . \text { of } e^{-} \mathrm{s} \text { involved }=2\right)$

Total number of $e^{-}$ s involved $=1+2=3$

(b) For $\mathrm{FeC}_{2} \mathrm{O}_{4}$

$\left.\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+} \quad \text { (No. of } e^{-\mathrm{s}} \text { involved }=1\right)$

$\left.\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow 2 \mathrm{CO}_{2} \text { (No. of } e^{-\mathrm{s}} \text { involved }=2\right)$

Total number of $e^{-}$ s involved $=1+2=3$

(c) For $\mathrm{Fe}\left(\mathrm{NO}_{2}\right)_{2}$

$\left.\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+} \quad \text { (No. of } e^{-} \text {s involved }=1\right)$

$\left.2 \mathrm{NO}_{2} \rightarrow 2 \mathrm{NO}_{3}^{-} \text {(No. of } e^{-} \text {s involved }=4\right)$

Total number of $e^{-s}$ involved $=1+4=5$

(d) For $FeSO_4$, $\left.\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+} \quad \text { (No. of } e^{-} \mathrm{s} \text { involved }=1\right)$

Total number of $e^-$ 's involved $= 1$

As $FeSO_4$, requires least number of electrons thus, it will require least amount of $\mathrm{KMnO}_{4}$