Using Gauss Law across the cylindrical Gaussian surface,\\\\(\\displaystyle\\int E.dS = \\dfrac{q_{enc}}{\\epsilon_0}\\\\)\\\\(\ herefore E\ imes 2\\pi r \ imes l = \\dfrac{\\lambda\ imes l}{\\epsilon_0}\\\\)\\\\(\ herefore E = \\dfrac{\\lambda}{2\\pi r\\epsilon_0}\\\\)Here \\\\(\\lambda\\\\) is the linear charge density on the wire.