500 mL of 0.2 M MnO _4^- solution in basic medium when mixed with...

First, we calculate the millimoles of the reactants:
Millimoles of $\text{MnO}_4^-=500 \text{ mL} \times 0.2 \text{ M} = 100 \text{ mmol}$
Millimoles of $\text{I}^-=500 \text{ mL} \times 1.5 \text{ M} = 750 \text{ mmol}$

In a basic medium, $\text{MnO}_4^-$ is reduced to $\text{MnO}_2$. The change in oxidation state of Mn is from $+7$ to $+4$, so its n-factor is $3$.
The problem explicitly states that iodide ions are oxidized to molecular iodine ($\text{I}_2$). The change in oxidation state for iodine is from $-1$ to $0$, so the n-factor for $\text{I}^-$ is $1$.

Equivalents of $\text{MnO}_4^-=100 \times 3 = 300 \text{ meq}$
Equivalents of $\text{I}^-=750 \times 1 = 750 \text{ meq}$

Since $\text{MnO}_4^-$ is the limiting reagent, the equivalents of $\text{I}_2$ produced will be equal to the equivalents of $\text{MnO}_4^-$ consumed.
Equivalents of $\text{I}_2$ formed = $300 \text{ meq}$

The liberated iodine is titrated with thiosulphate ($\text{S}_2\text{O}_3^{2-}$):
$\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$

For thiosulphate, the n-factor is $1$ (as $2$ moles of $\text{S}_2\text{O}_3^{2-}$ lose $2$ moles of electrons).
From the law of equivalence:
Equivalents of $\text{S}_2\text{O}_3^{2-}$ = Equivalents of $\text{I}_2$
$\text{Molarity} \times \text{Volume (in mL)} \times \text{n-factor} = 300$
$x \times 300 \times 1 = 300$
$x = 1$

Answer: $1$