Answer
$E = 1.31\times 10^{-6}~N/C$
Work Step by Step
If $x = 2.00~mm$, then by symmetry, the electric field due to the charge between $x = 2.00~mm$ and $x = 4.70~mm$ cancels out with the charge between $x = -0.70~mm$ and $x = 2.00~mm$
We can think of the slab as a having a surface charge density of $\sigma$ and we can consider the charge between $x = -4.70~mm$ and $x = -0.70~mm$. Note that this interval has a width of $4.00~mm$
We can find the value of $\sigma$:
$\sigma = (\rho)~(4.00~mm)$
$\sigma = (5.80\times 10^{-15}~C/m^3)(4.00\times 10^{-3}~m)$
$\sigma = 2.32\times 10^{-17}~C/m^2$
We can find the magnitude of the electric field:
$E = \frac{\sigma}{2\epsilon_0}$
$E = \frac{2.32\times 10^{-17}~C/m^2}{(2)(8.854\times 10^{-12}~F/m)}$
$E = 1.31\times 10^{-6}~N/C$
