Number of mtextlli equivalents of $\text{M}\text{n}{\text{O}}_4^{-}=0.02\times 3\times 250=15$

Number of mtextlli equivalents of ${\text{l}}_2=0.1\times 1\times 250=25$

Thus, here $\text{M}\text{n}{\text{O}}_4^{-}$ is limtextting reagent.

$\therefore$ Number of mtextlli equivalents of ${\text{l}}_2$ formed = Number of mtextlli equivalent of $\text{M}\text{n}{\text{O}}_4^{-}=15$

Or number of equivalent of ${\text{l}}_2$ formed $=\frac{15}{1000}=0.015$

$\therefore$ Number of moles of ${\text{l}}_2$ formed $=\frac{0.015}{2}$

$=0.0075$