In alkaline medium permanganate is reduced to $\mathrm{MnO}{}_2$. The number of electrons gained by one $\mathrm{MnO}{}_4{}^{-}$ ion in this half-reaction is

Mn changes from +7 in $\mathrm{MnO}{}_4{}^{-}$ to +4 in $\mathrm{MnO}{}_2$, a change of 3 units. So 3 electrons are gained: $\mathrm{MnO}{}_4{}^{-}+2\mathrm{H}{}_2\mathrm{O}+3\mathrm{e}{}^{-}\to \mathrm{MnO}{}_2+4\mathrm{OH}{}^{-}$. The answer is 3 (5 applies to acidic medium, 1 to strongly alkaline medium giving $\mathrm{MnO}{}_4{}^{2-}$).