Answer
$q_B = -3~q_0$
Work Step by Step
We can draw a Gaussian cylinder through point 2.
According to Gauss' law: $~~\epsilon_0~\Phi = q_{enc}$
Then:
$\epsilon_0~E~(2\pi~r~h) = q_{enc}$
$E = \frac{q_{enc}}{2~\epsilon_0~\pi~r~h}$
Note that $q_{enc} = q_A+q_B$
We can find the required value of $q_B$:
$q_{enc} = q_A+q_B = 0$
$q_B = -q_A$
$q_B = -3~q_0$
