As potential difference between B & D is zero therefore it is a balanced wheatstone bridge. So, we can write

$$\begin{gathered} \Rightarrow \frac{R_1}{R_2}=\frac{R_3}{R_4} \\ \Rightarrow \frac{\left({\displaystyle \frac{6\times 3}{6+3}}\right)}{1+2}=\frac{\left({\displaystyle \frac{x\times 1}{x+1}}\right)}{x} \end{gathered}$$

$\Rightarrow \frac{2}{3}=\frac{1}{x+1}$

$\Rightarrow x=0.5=\frac{1}{2}$

Hence, required $n=2$.