Why this redox equation occurs on basic medium?
You are given a net ionic equation, which is the reduced form of a complete ionic equation: that's why you see neither $\ce{H2O}$ nor $\ce{OH-}$.
You can deduce that the medium should be slightly alkaline a priori only if you:


either know that manganate $\ce{MnO4^2-}$ formed in alkaline medium hydrolyses when there is a lack of reducing agent (here, sulfite $\ce{SO3^2-}$) to manganese(IV) oxide $\ce{MnO2}$ and that $\ce{MnO2}$ is also formed near neutral $\mathrm{pH}$ values;
or you compose the redox reaction as you were asked to and notice that two $\ce{OH-}$ 
are produced:


$$
\begin{align}
\ce{\overset{+7}{Mn}O4- + 2 H2O + 3 e- &amp;→ \overset{+4}{Mn}O2 + 4 OH-} &amp; |\cdot 2 \tag{red}\\
\ce{\overset{+4}{S}O3- + 2 OH- &amp;→ \overset{+6}{S}O4^2- + H2O + 2 e-} &amp; |\cdot 3 \tag{ox}\\
\hline
\ce{2\overset{+7}{Mn}O4- + 3\overset{+4}{S}O3- + H2O &amp;→ 2\overset{+4}{Mn}O2 + 3\overset{+6}{S}O4^2- + 2 OH-} \tag{redox}
\end{align}
$$

As for why your problem states the redox reaction takes place in basic medium, it's just an extra hint assisting you in balancing the oxidation and reduction half-reactions (number of oxygens in oxides and oxoanions, for instance), but you could've balance the equation all right even if there were no information about the alkalinity of the medium.