A long cylindrical volume contains a uniformly distributed charge of density ρ   C   m - 3 .…

Given here, Volume charge density $=\rho \mathrm{C} {\mathrm{m}}^{-3}$.

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,

$\int EdS\cos 0=\frac{q}{{\epsilon }_0}$

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

 $\Rightarrow E\left(2\pi xh\right)=\frac{\rho \times \pi x^{2}h}{{\epsilon }_0}$

$\Rightarrow E=\frac{\rho x}{2{\epsilon }_0}$  

$\Rightarrow E=\frac{\rho }{2{\epsilon }_0}\times \frac{2{\epsilon }_0}{\rho }=1$

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