The given sequence is $\sqrt{3}, 3,3 \sqrt{3,}, \ldots \ldots$

$a=\sqrt{3}$ and $r=\frac{3}{\sqrt{3}}=\sqrt{3}$

Let the $n^{\text {th }}$ term of the given sequence be $729 .$

$a_{n}=a r^{n-1}$

$\therefore a r^{n-1}=729$

$\Rightarrow(\sqrt{3})(\sqrt{3})^{n-1}=729$

$\Rightarrow(3)^{1 / 2}(3)^{\frac{n-1}{2}}=(3)^{6}$

$\Rightarrow(3)^{\frac{1}{2}+\frac{n-1}{2}}=(3)^{6}$

$\therefore \frac{1}{2}+\frac{n-1}{2}=6$

$\Rightarrow \frac{1+n-1}{2}=6$

$\Rightarrow n=12$

Thus, the $12^{\text {th }}$ term of the given sequence is $729 .$