The change in oxidation state of A when,

$A^{n+}$ is oxidised to $AO_3^{-}$ in acidic medium is form $+n$ to $+n \;+0 \; (+5)$ ie the change is equal to $(5-n)$

The number of electrons added to reduced

$1.61 \times 10^{-3}$ moles of $MnO^{-}_4$ to $Mn^{2+}$

(o.s of Mn changes from $+7$ to $+2=1.61 \times 10^{-3} \times 5)$

$\therefore 1.61 \times 10^{-3} \times 5 =2.68 \times 10^{-3}(5-n)$

or $(5-n) =\large\frac{1.61 \times 10^{-3} \times 5}{2.68 \times 10^{-3}}$

or $5-n=3=>n=2$

Hence b is the correct answer.