\\\\(\ extbf{Step 1: Electric field due to infinite line charge}\\\\) Electric field due to infinite line charge is given by: \\\\(E = \\dfrac{\\lambda}{2\\pi \\in_0 d}\\\\) \\\\(\\Rightarrow \\lambda = 2\\pi \\in_0 E \\ d\\\\) \\\\(....(1)\\\\)Where \\\\(\\lambda\\\\) is the linear charge density and \\\\(d\\\\) is the perpendicular distance of point from the infinite line charge.\\\\(\ extbf{Step 2: Substituting the values }\\\\) \\\\(\\because \\dfrac{1}{4\\pi \\in_0} = 9 \ imes 10^9 Nm^2C^{-2}\\\\) \\\\(\\Rightarrow 2\\pi \\in_0 = \\dfrac{1}{18\ imes 10^9 Nm^2C^{-2}}\\\\)From Eq \\\\((1): \\\\) \\\\(\\lambda = \\dfrac{1 \ imes 9 \ imes 10^4\\ NC^{-1}\ imes 0.02m}{18\ imes 10^9\\ N m^2 C^{-2} }= 10^{-7}\\ Cm^{-1}\\\\)