first, check your data booklet for the relevant half-equations
(1) MnO4- (aq) + 8H+ (aq) +5e -> 4H2O (l) + Mn2+ (aq)
(2) 2Br- (aq) -> Br2 (aq) + 2e
Equation (1) uses up 5e while Equation (2) uses up 2e. Since the lowest common multiple between 5 and 2 is ten, we ought to apply the mole-multiplier of 2 and 5 respectively on each equation, such that they involve 10e each.
The modfied half equations would be:
2MnO4- (aq) + 16H+ (aq) +10e -> 8H2O (l) + 2Mn2+ (aq)
10Br- (aq) -> 5Br2 (aq) + 10e
Next, combine both modified half-equations to obtain the final chemical equation, ie:
2MnO4- (aq) + 16H+ (aq) + 10Br- (aq) -> 8H2O (l) + 2Mn2+ (aq) + 5Br2 (aq)