The electric field generated by an infinite plane uniformly charged is
[tex]E= \frac{\sigma}{2\epsilon _0} [/tex]
where [tex]\sigma[/tex] is the charge density of the plane, while [tex]\epsilon _0 = 8.85\cdot 10^{-12} F/m[/tex] is the vacuum permittivity.
Let's calculate the electric field of the first (xy) plane:
[tex]E_1 = \frac{\sigma_1}{2 \epsilon _0}= \frac{8\cdot 10^{-9} C/m^2}{2 \cdot
8.85\cdot 10^{-12} F/m}= 452.0 V/m[/tex]
And now the electric field due to the second (z=2.0 m) plane:
[tex]E_2 =\frac{\sigma_2}{2 \epsilon _0}= \frac{3\cdot 10^{-9} C/m^2}{2 \cdot 8.85\cdot 10^{-12} F/m}= 169.5 V/m[/tex]
The electric field of the two planes does not depend on the distance. Also, the charges on the two planes have same sign, so at z=3.0 m (and at every point with z>2.0m) the two fields point into the same direction and the total electric field is simply the sum of the two fields:
[tex]E_{tot} = E_1 + E_2 =621.5 V/m[/tex]
