Answer
The net electric field is zero at the point $~~x = -3.3~cm$
Work Step by Step
We can find the charge on Shell 1:
$q_1 = \sigma_1~4\pi~r_1^2$
$q_1 = (+4.0\times 10^{-6}~C/m^2)~(4\pi)~(0.0050~m)^2$
$q_1 = +1.2566 \times 10^{-9}~C$
We can find the charge on Shell 2:
$q_2 = \sigma_2~4\pi~r_2^2$
$q_2 = (-2.0\times 10^{-6}~C/m^2)~(4\pi)~(0.020~m)^2$
$q_2 = -1.0053 \times 10^{-8}~C$
The electric field due to the positive charge on Shell 1 is directed away from Shell 1, and the electric field due to the negative charge on Shell 2 is directed toward Shell 2.
Since the magnitude of the charge on Shell 2 is larger than the magnitude of the charge on Shell 1, the point where the net electric field is zero will be located to the left of Shell 1.
Let the distance to this point from Shell 1 be $d$.
We can find $d$:
$\frac{q_1}{4\pi~\epsilon_0~d^2}+\frac{q_2}{4\pi~\epsilon_0~(d+L)^2} = 0$
$\frac{q_1}{4\pi~\epsilon_0~d^2} = -\frac{q_2}{4\pi~\epsilon_0~(d+L)^2}$
$\frac{+1.2566 \times 10^{-9}~C}{4\pi~\epsilon_0~d^2} = -\frac{-1.0053 \times 10^{-8}~C}{4\pi~\epsilon_0~(d+L)^2}$
$\frac{1}{d^2} = \frac{8.0}{(d+L)^2}$
$(d+L)^2 = 8.0~d^2$
$d+L = \sqrt{8.0}~d$
$\sqrt{8.0}~d - d = L$
$d = \frac{L}{\sqrt{8.0} - 1}$
$d = \frac{6.0~cm}{\sqrt{8.0} - 1}$
$d = 3.3~cm$
Since this distance is to the left of Shell 1, the net electric field is zero at the point $~~x = -3.3~cm$
