It requires  of $40 \mathrm{mL}$ to $1M\mathit{ }{\mathrm{Ce}}^{4+}$titrate $20 \mathrm{mL}$ of  $1M$ ${\mathrm{Sn}}^{2+}$to ${\mathrm{Sn}}^{4+}$. What is the oxidation state of the cerium in the product?

Number of equivalents of $C{e}^{4+}=$ Number of equivalents of ${\mathrm{Sn}}^{2+}$

$40.05\times 1\times n$ factor $=20\times 1\times 2$

$\Rightarrow \mathrm{n}$ factor of ${\mathrm{Ce}}^{4+}=1$

Therefore oxidation state change will be

Hence, the correct answer is 3.