Answer
$$\therefore E= \frac{1}{4\pi \epsilon_0}\frac{2\lambda}{s}$$
Work Step by Step
Consider a Gaussian cylindrical surface of radius s and length l.
Let the wire passes through the axis of the cylinder.
By Guass Law,
$$\oint \vec{E}\cdot \vec{da} = \frac{Q_{enc}}{\epsilon_0}$$
$\because $ Electric field through the plane surfaces of the cylinder is zero.
$$\therefore E\times 2\pi s l= \frac{\lambda\times l}{\epsilon_0}$$
$$\therefore E= \frac{1}{2\pi \epsilon_0}\frac{\lambda}{s}$$
$$\therefore E= \frac{1}{4\pi \epsilon_0}\frac{2\lambda}{s}$$
