Total charge required for oxidation of 2 moles Mn₃O₄ into MnO₄ ^2- in the presence of alkaline medium is:

  Oxidation of Mn₃O₄ to MnO₄^2- takes place in following steps:

$$\begin{gathered} {\mathrm{2Mn}}_3{O}_4{+\; 2OH}^{-}\to {\mathrm{3Mn}}_2{O}_3+\; 2e^{} +H_{2}O \\ {\mathrm{3Mn}}_2{O}_3{+\; 6OH}^{-}\to {\mathrm{6MnO}}_2+\; 6e +H_{2}O \\ {\mathrm{6MnO}}_2{+\; 24OH}^{-}\to {{\mathrm{6MnO}}_4}^{\mathrm{2-}}+\; 12H_{2}O +12e \end{gathered}$$

Total moles of electrons = 20
So, total charge required = 20 x 96500 F