How to balance the disproportionation reaction of manganate anion in basic conditions?
You are referring to the disproportionation reaction of $\ce{Mn(VI)}$ to $\ce{Mn(IV)}$ and $\ce{Mn(VII)}$, which is:

$\ce{MnO4^{2-}(aq) -&gt; MnO2 (s) + MnO4^{-}(aq)}$

Reduction: $\ce{MnO4^{2-} + 2e^{-} + 2H2O -&gt; MnO2 + 4OH-}$

Oxidation: $\ce{2(MnO4^{2-} -&gt; MnO4^{-} + 1e^{-})}$ = $\ce{2MnO4^{2-} -&gt; 2MnO4^{-} + 2e^{-} }$

Combine both, and the final balanced equation is:

$\ce{3MnO4^{2-} + 2H2O -&gt; MnO2 + 2MnO4^{-} + 4OH-}$