Given here, Volume charge density =ρCm−3.
Consider a Gaussian cylindrical surface of height h and radius x.
Let q charge is enclosed by the cylinder of radius x.

From Gauss's law,
∫EdScos0=ϵ0q
Here, q=volume charge density×volume of imaginary cylinder.
⇒E(2πxh)=ϵ0ρ×πx2h
⇒E=2ϵ0ρx
⇒E=2ϵ0ρ×ρ2ϵ0=1
Thus, electric field is 1Vm−1.