Answer
At $x = 8.0~cm,~~$ the net electric field is zero.
Work Step by Step
The electric field due to the positively charged line will be directed away from that line.
The electric field due to the negatively charged line will be directed toward that line.
Since the positively charged line has a greater magnitude of charge and the electric field magnitude decreases as the distance from the charge increases, the point where $E_{net} = 0$ will be to the right of the negatively charged line.
We can add the electric fields due to each charged line in order to find $x$:
$E_{net} = \frac{\lambda_1}{2\pi~\epsilon_0~r_1}+\frac{\lambda_2}{2\pi~\epsilon_0~r_2} = 0$
$\frac{6.0~\mu C/m}{2\pi~\epsilon_0~(x+L/2)}-\frac{2.0~\mu C/m}{2\pi~\epsilon_0~(x-L/2)} = 0$
$\frac{6.0~\mu C/m}{2\pi~\epsilon_0~(x+L/2)} = \frac{2.0~\mu C/m}{2\pi~\epsilon_0~(x-L/2)}$
$\frac{3.0}{x+L/2} = \frac{1.0}{x-L/2}$
$3.0x-3.0L/2 = x+L/2$
$2.0x = 2L$
$x = L$
$x = 8.0~cm$
At $x = 8.0~cm,~~$ the net electric field is zero.
