Consider a gaussian cylinder of radius r and length L. The charge enclosed by the cylinder is rho (pi r2 L) and the flux through the tube wall is E (2 pi r L). There is no flux contribution from the endcaps. By Gauss's law then the electric field is proportional to the distance from the center and to the charge density. E = (rho r) / (2 ep0), pointing inward.
Outside the cylinder a "line of charge" result is obtained.
Thanks to Andrew Field for pointing out a correction to the statement on charge enclosed and to Leslie Rossman for pointing out a missing division sign.