KMnO_4 reacts with KI , in basic medium to form I_2 and MnO_2 ....

$$\begin{array}{r}\lceil\text {-factor }=1 \longrightarrow \\ \mathrm{MnO}_4^{-}+\mathrm{I}^{-} \rightleftharpoons \mathrm{MnO}_2+\mathrm{I}_2 \\ \bigsqcup_{n \text {-factor }=3}^{\rightleftharpoons}\end{array}$$Number of milli equivalents of $\mathrm{MnO}_4^{-}=0.02 \times 3 \times 250=15$ Number of milli equivalents of $\mathrm{I}_2=0.1 \times 1 \times 250=25$ Thus, here $\mathrm{MnO}_4^{-}$is limiting reagent. $\therefore$ Number of milli equivalents of $\mathrm{I}_2$ formed = Number of milli equivalent of $\mathrm{MnO}_4^{-}=15$ or number of equivalent of $\mathrm{I}_2$ formed =\frac{15}{1000}=$0.$015\begin{aligned} \therefore \text { Number of moles of } \mathrm{I}}_{2} \text { formed } & =\frac{0.015}{2}\\ & =0.0075 \end{aligned}$