Hello candidate,

First take the assumption- Let us draw a gaussian cylinder of length l and radius r across the line of charge having density λc/m.

At cylindrical part of the surface electric field  vector E is normal to the surface at every point and its magnitude is constant. Therefore flux through the Gaussian surface. = Flux through the curved cylindrical part of the surface. = E x 2πrl and hence, the expression for electric field is given by lambda/ 2π£r.

Hope It was helpful!!