1. We need to balance:

$2\mathrm{MnO_4^-} + b\mathrm{I^-} + c\mathrm{H_2O} \rightarrow x\mathrm{I_2} + y\mathrm{MnO_2} + z\mathrm{OH^-}$

Since there are already $2$ permanganate ions on the left, let us balance Mn first.

2. Balance manganese:

Each $\mathrm{MnO_4^-}$ gives one Mn atom, so total Mn on left = $2$. Thus,

$y = 2$

So the equation becomes

$2\mathrm{MnO_4^-} + b\mathrm{I^-} + c\mathrm{H_2O} \rightarrow x\mathrm{I_2} + 2\mathrm{MnO_2} + z\mathrm{OH^-}$

3. This is a redox reaction in basic medium, so use half-reactions.

Oxidation half-reaction

Iodide is oxidized to iodine:

$2\mathrm{I^-} \rightarrow \mathrm{I_2} + 2e^-$

Reduction half-reaction

Permanganate is reduced to $\mathrm{MnO_2}$ in basic medium. Start in acidic form:

$\mathrm{MnO_4^-} + 4\mathrm{H^+} + 3e^- \rightarrow \mathrm{MnO_2} + 2\mathrm{H_2O}$

Convert to basic medium by adding $4\mathrm{OH^-}$ to both sides:

$\mathrm{MnO_4^-} + 4\mathrm{H^+} + 4\mathrm{OH^-} + 3e^- \rightarrow \mathrm{MnO_2} + 2\mathrm{H_2O} + 4\mathrm{OH^-}$

Since $\mathrm{H^+} + \mathrm{OH^-} = \mathrm{H_2O}$,

$\mathrm{MnO_4^-} + 4\mathrm{H_2O} + 3e^- \rightarrow \mathrm{MnO_2} + 2\mathrm{H_2O} + 4\mathrm{OH^-}$

Cancel $2\mathrm{H_2O}$ from both sides:

$\mathrm{MnO_4^-} + 2\mathrm{H_2O} + 3e^- \rightarrow \mathrm{MnO_2} + 4\mathrm{OH^-}$

4. Since there are $2\mathrm{MnO_4^-}$, multiply the reduction half-reaction by $2$:

$2\mathrm{MnO_4^-} + 4\mathrm{H_2O} + 6e^- \rightarrow 2\mathrm{MnO_2} + 8\mathrm{OH^-}$

5. To balance electrons, multiply the oxidation half-reaction by $3$:

$6\mathrm{I^-} \rightarrow 3\mathrm{I_2} + 6e^-$

6. Add the two half-reactions:

$2\mathrm{MnO_4^-} + 4\mathrm{H_2O} + 6\mathrm{I^-} \rightarrow 2\mathrm{MnO_2} + 8\mathrm{OH^-} + 3\mathrm{I_2}$

Rewriting in the given form:

$2\mathrm{MnO_4^-} + 6\mathrm{I^-} + 4\mathrm{H_2O} \rightarrow 3\mathrm{I_2} + 2\mathrm{MnO_2} + 8\mathrm{OH^-}$

7. Therefore,

$z = 8$

8. Comparison with stored correct answer:

Stored correct answer = $8$

Our derived answer = $8$

Hence, the answer agrees.