The sum of the rational terms of ${\left(2^{1/5}+\sqrt{3}\right)}^{20}$is

$T_{r+1,}$the $\left(r+1\right)$th term in the expansion of ${\left(2^{1/5}+\sqrt{3}\right)}^{20}$is

      $T_{r+1}{=}^{20}{C}_r{\left(2^{1/5}\right)}^{20-r}{\left(3^{1/2}\right)}^r{=}^{20}{C}_r{2}^{4-r/5}(3{)}^{r/2}$
As $2$ and $3$ are relatively prime $2^{4-r/5}$ and $3^{r/2}$ will be rational if both $4-r/5$ and $r/2$ are integers, i.e. if $r/5$ and $r/2$ are integers. This is possible if
                 $r=0,10,20.$
Thus, the sum of rational terms is
${ }^{20}{C}_0\left(2^4\right)\left(3^0\right){+}^{20}{C}_{10}\left(2^2\right)\left(3^5\right){+}^{20}{C}_{20}\left(2^0\right)\left(3^{10}\right)$
As this is more than $71, 85$ or $97$, answer is (4).