The electric field at distance ‘r’ from infinite line of charge (having linear charge density $\mathrm{\lambda }$ ) is __

On considering a Gaussian surface (cylinder in this case) of radius r and length l. 
Then according to the Gauss's Law of electrostatics:

$\int \overrightarrow{\mathrm{E}}·\overrightarrow{\mathrm{dA}}=\frac{{\text{ q}}_{\text{enclosed}}}{{\mathrm{ϵ}}_0}$

From the symmetry of the surface, we can easily say that the electric field would only be along the radial direction. And the amount of charge enclosed by this cylinder is charge enclosed =λ×l and the surface area of the cylinder A=2πrl. 

$\Rightarrow \mathrm{E}\times 2\mathrm{\pi r}l=\frac{\mathrm{\lambda }l}{{\mathrm{ϵ}}_0}$

$\mathrm{E}=\left(\frac{1}{2{\mathrm{\pi ϵ}}_0}\right)\frac{\mathrm{\lambda }}{\mathrm{r}}$