first, check your data booklet for the relevant half-equations

(1) MnO_{4}^{-} (aq) + 8H^{+} (aq) +5e -> 4H_{2}O (l) + Mn^{2+} (aq)

(2) 2Br^{-} (aq) -> Br_{2} (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:

2MnO_{4}^{-} (aq) + 16H^{+} (aq) +10e -> 8H_{2}O (l) + 2Mn^{2+} (aq)

10Br^{-} (aq) -> 5Br_{2} (aq) + 10e

Next, combine both modified half-equations to obtain the final chemical equation, ie:

2MnO_{4}^{-} (aq) + 16H^{+} (aq) + 10Br^{-} (aq) -> 8H_{2}O (l) + 2Mn^{2+} (aq) + 5Br_{2} (aq)