Answer
There is 25% increase in the surface area after the reaction
Work Step by Step
The surface area $a$ of a nucleus is given by
$$
a \simeq 4 \pi R^{2} \simeq 4 \pi\left(R_{0} A^{1 / 3}\right)^{2} \propto A^{2 / 3}
$$
We know that a $^{236}$U nucleus breaks to form $^{140}$Xe and $^{96}$Sr.
So the final area will comprise of combined area of Xe & Sr.
Thus, the fractional change in surface area is
$$
\frac{\Delta a}{a_{i}}=\frac{a_{f}-a_{i}}{a_{i}}=\frac{(140)^{2 / 3}+(96)^{2 / 3}}{(236)^{2 / 3}}-1=+0.25
$$
$\therefore$ there is 25% increase in the surface area after the reaction
