Answer
$E = (-2.8\times 10^4~N/C)~\hat{i}$
Work Step by Step
Since $x = 2.0~cm$ is inside Shell 1, the electric field due to the charge on Shell 1 is zero.
We can find the charge on Shell 2:
$q = 4\pi~r^2~\sigma$
$q = (4\pi)~(0.020~m)^2~(4.0\times 10^{-6}~C/m^2)$
$q = 2.01\times 10^{-8}~C$
We can find the electric field at $x = 2.0~cm$:
$E = \frac{q}{4\pi~\epsilon_0~d^2}$
$E = \frac{2.01\times 10^{-8}~C}{(4\pi)~(8.854\times 10^{-12}~F/m)~(0.080~m)^2}$
$E = 2.8\times 10^4~N/C$
Note that since the charge on Shell 2 is positive, the net electric field at $x = 2.0~cm$ is in the negative direction of the x axis.
We can express the net electric field in unit-vector notation:
$E = (-2.8\times 10^4~N/C)~\hat{i}$
