How many Coulomb's of electricity are required for the reduction of 1 mole of $Mn{O}_4^{-}toM{n}^{2+}$

The correct option is C $4.83\times {10}^{5}C$
First let's write down the reaction.
$M{n}^{7+}+5e^{-}\to M{n}^{2+}$
In this reduction reaction, you can see that for 1 mole of $M{n}^{7+}$ we will need 5 mole of electrons for conversion to $M{n}^{2+}$
Therefore, we need $5\times 96500C=4.83\times {10}^{5}Cofcharge$