In the following reaction

${\mathrm{Fe}}_3{\mathrm{O}}_4+{\mathrm{MnO}}_4^{-}⟶{\mathrm{Fe}}_2{\mathrm{O}}_3+{\mathrm{MnO}}_2$

the number of moles of OH^- liberated as per the balanced equation in the basic medium is

$\therefore {\mathrm{Fe}}_3{\mathrm{O}}_4\text{ and }{\mathrm{MnO}}_4^{-}\text{react in }6:2\text{ mole ratio in the balanced }$chemical equation 

$6{\mathrm{Fe}}_3{\mathrm{O}}_4+2{\mathrm{MnO}}_4^{-}+{\mathrm{H}}_2\mathrm{O}⟶9{\mathrm{Fe}}_2{\mathrm{O}}_3+2{\mathrm{MnO}}_2+{\mathrm{OH}}^{-}$

by balancing the residual charge with OH^-

$6{\mathrm{Fe}}_3{\mathrm{O}}_4+2{\mathrm{MnO}}_4^{-}+{\mathrm{H}}_2\mathrm{O}⟶9{\mathrm{Fe}}_2{\mathrm{O}}_3+2{\mathrm{MnO}}_2+2{\mathrm{OH}}^{-}$