Solution:

Given that,
The sum of GP $=39+13 \sqrt{3}$
Where, $a=\sqrt{3}, r=3 / \sqrt{3}=\sqrt{3}, n=?$
Using the formula,
The sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$
$39+13 \sqrt{3}=\sqrt{3}\left(\sqrt{3^{n}}-1\right) /(\sqrt{3}-1)$
$(39+13 \sqrt{3})(\sqrt{3}-1)=\sqrt{3}\left(\sqrt{3}^{n}-1\right)$
On simplifying we obtain
$\begin{array}{l}
39 \sqrt{3}-39+13(3)-13 \sqrt{3}=\sqrt{3}\left(\sqrt{3}^{n}-1\right) \\
39 \sqrt{3}-39+39-13 \sqrt{3}=\sqrt{3}\left(\sqrt{3}^{n}-1\right) \\
39 \sqrt{3}-39+39-13 \sqrt{3}=\sqrt{3}^{n+1}-\sqrt{3} \\
26 \sqrt{3}+\sqrt{3}=\sqrt{3}^{n+1} \\
27 \sqrt{3}=\sqrt{3}^{n+1} \\
\sqrt{3}^{6} \sqrt{3}=\sqrt{3}^{n+1} \\
6+1=n+1 \\
7=n+1 \\
7-1=n \\
6=n
\end{array}$
As a result, 6 terms are required to make a sum of $39+13 \sqrt{3}$