It's hard to pinpoint what the problem here is since I don't have the mentioned book to compare an answer with.
The simplest way is write down half-reactions for reduction (red) and oxidation (ox) processes once you've assigned oxidation numbers (denoted above the symbols of the elements which are participating in a redox reaction), then balance the number of the transferred electrons and the algebraic sum of both half-reactions will yield in a complete balanced redox reaction:
$$
\begin{align}
\ce{\overset{+7}{Mn}O4- + 2 H2O + 3 e- &→ \overset{+4}{Mn}O2 + 4 OH-} &\quad |\cdot 2 \tag{red}\\
\ce{\overset{-1}{Br}^- + 6 OH- &→ \overset{+5}{Br}O3- + 3 H2O + 6 e-} &\quad \tag{ox}\\
\hline
\ce{2 MnO4- + Br- + H2O &→ 2 MnO2 + BrO3- + 2 OH-} \tag{redox}
\end{align}
$$
As a sanity check, you can see that there are no protons $\ce{H+}$ participating in either of the reactions, and the resulting medium is indeed basic (slight excess of $\ce{OH-}$), which is in a good agreement with the conditions mentioned in the problem.
