$15 mL$ of aqueous solution of ${Fe}^{2+}$ in acidic medium completely reacted with $20 mL$ of $0.03M$ aqueous ${Cr}_2{O}_7{ }^{2-}$. The molarity of the ${Fe}^{2+}$ solution is ${10}^{-2}\mathrm{M}$. (Round off to the Nearest Integer).

$M_{eq}$ of ${Fe}^{2+}=M_{eq}$ of ${Cr}_2{O}_7^{2-}$

$M\times 15\times n-$ factor $=0.03\times 20\times n-$ factor

In the reaction, ${Fe}^{2+}$ is oxidised to ${Fe}^{3+}$ hence, the $n$-factor is 1.

Also, ${Cr}_2{O}_7^{2-}$ is reduced to $2{Cr}^{3+}$. Thus, 6 electrons are involved. Hence, its $n$-factor is 6.

Molarity $=0.24M\Rightarrow 24\times {10}^{-2}M$

Hence the correct answer is 24.