Answer
$E = 3.08\times 10^{-6}~N/C$
Work Step by Step
Note that $x = 4.70~mm$ is at a point that is on the surface of the slab of charge.
We can think of the slab as a having a surface charge density of $\sigma$.
We can find the value of $\sigma$:
$\sigma = \rho~d$
$\sigma = (5.80\times 10^{-15}~C/m^3)(9.40\times 10^{-3}~m)$
$\sigma = 5.452\times 10^{-17}~C/m^2$
We can find the magnitude of the electric field:
$E = \frac{\sigma}{2\epsilon_0}$
$E = \frac{5.452\times 10^{-17}~C/m^2}{(2)(8.854\times 10^{-12}~F/m)}$
$E = 3.08\times 10^{-6}~N/C$
