Imagine , a cylindrical gaussian surface of length L and radius a.

$\begin{array}{l}\tan \left(\angle APO\right)=\frac{\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$\sqrt{3}a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{1}{\sqrt{3}}\text{}\\ \Rightarrow \angle APO={30}^0\\ \text{}Now , \angle APD=2\angle APO={60}^0\end{array}$

Given rectangular surface ABCD subscribes an angle  ${60}^0$ at the infinity long uniform line of charge.

The total angle subscribe by cylinder is ${360}^0$ and charge enclosed is $\lambda L$.

Flux through the rectangle is same as the flux through ${\frac{1}{6}}^{th}$ of the hexagon since it subtends ${60}^0$ at P and field lines are symmetrically radially outwards.

$\begin{array}{l}{\varphi }_{cyl}=\frac{q}{{\in }_0}\text{}\\ \therefore {\varphi }_{ABCD}=\frac{q}{6{\in }_0}=\frac{\lambda L}{6{\in }_0}\text{}\\ \Rightarrow n=6\end{array}$