Answer
$E = 5.4~N/C$
Work Step by Step
We can draw a Gaussian cylinder with two end faces of area $A$ that goes through the slab between $x = -4.0~cm$ and $x = 4.0~cm$.
We can find the electric field at $x = 4.0~cm$:
$\epsilon_0~\Phi = q_{enc}$
$\epsilon_0~2A~E = \rho A~d$
$E = \frac{\rho ~d}{2~\epsilon_0}$
$E = \frac{(1.2\times 10^{-9}~C/m^3)(0.080~m)}{(2)~(8.854\times 10^{-12}~F/m)}$
$E = 5.4~N/C$
