In acidic solution, $E^{\circ }(\mathrm{MnO}{}_4{}^{-}/\mathrm{MnO}{}_2)=1.70\mathrm{V}$ (3 electrons) and $E^{\circ }(\mathrm{MnO}{}_2/\mathrm{Mn}{}^{2+})=1.23\mathrm{V}$ (2 electrons). $E^{\circ }(\mathrm{MnO}{}_4{}^{-}/\mathrm{Mn}{}^{2+})$ is

Add the Gibbs energies: $-5F{E}^{\circ }=-3F(1.70)-2F(1.23)$. So $E^{\circ }=\frac{3(1.70)+2(1.23)}{5}=\frac{5.10+2.46}{5}=1.512\approx 1.51\mathrm{V}$. The result is the electron-weighted mean of the two potentials. Option B is the simple average, option D swaps the weights, option C adds the potentials.