The given sequence is $\sqrt 3,3,3\sqrt 3.........$

This sequence is a G.P.since the ratio of any two successive terms is same.

$1^{st}\:term$ of the G.P.$=a=\sqrt 3$ and common ratio$=r=\large\frac{3}{\sqrt 3}$$=\sqrt 3$

We know that $t_n=a.r^{n-1}$

Also given that $t_n=729$

$\Rightarrow\:729=a.r^{n-1}=\sqrt 3.(\sqrt 3)^{n-1}=(\sqrt 3)^{n-1+1}=(\sqrt 3)^n$

$729=3^6$ and $\sqrt 3=3^{1/2}$

$\Rightarrow\:3^6=3^{n/2}$

Since the base on either side is equal (=3) the powe on either side is also equal.

$\Rightarrow\:6=\large\frac{n}{2}$ or $n=12$

$i.e., 12^{th}$ term of the given sequence is $729$