A long cylindrical volume contains a uniformly distributed charge...

Given here, Volume charge density $= ρC{m}^{- 3}$.

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,

$∫ E d Scos0 = \dfrac{q}{{ε}_{0}}$

Here, $q =$volume charge density$\times$volume of imaginary cylinder.

$⇒ E\left(2\pixh\right) = \dfrac{ρ \times \pi{x}^{2}h}{{ε}_{0}}$

$⇒ E = \dfrac{ρx}{2{ε}_{0}}$

$⇒ E = \dfrac{ρ}{2{ε}_{0}} \times \dfrac{2{ε}_{0}}{ρ} = 1$

Thus, electric field is $1V{m}^{- 1}$.