Answer
See the detailed answer below.
Work Step by Step
We know that the electric field exerted by an infinite line charge at a distance $r$ from it is given by
$$E_{line}=\dfrac{1}{4\pi \epsilon_0}\;\dfrac{2|\lambda|}{r}$$
Now we need to draw the electric field direction as seen below.
From the geometry of this graph, we can see that
$\bullet$ $\theta_L=\theta_R=\theta$
$\bullet$ $r_L=r_R=\sqrt{(d/2)^2+y^2}$
$\bullet$ $\sin\theta=\dfrac{d/2}{r}=\dfrac{d}{ 2\sqrt{(d/2)^2+y^2}}$
$\bullet$ $\cos\theta=\dfrac{y}{r}=\dfrac{y}{\sqrt{(d/2)^2+y^2}}$
It is obvious, from the figure below, that the $y$ and $z$-components of the electric fields from both lines cancel each other and that the net electric field is in the $x$-direction.
Hence,
$$E_{net}=E_L+E_R$$
$$E_{net}= \dfrac{2 \lambda }{4\pi \epsilon_0 r}(-\sin\theta\;\hat i+\cos\theta\;\hat j)+\dfrac{2 \lambda }{4\pi \epsilon_0 r}(-\sin\theta\;\hat i-\cos\theta\;\hat j)$$
$$E_{net}= \dfrac{-4 \lambda }{4\pi \epsilon_0 r} \sin\theta\;\hat i$$
plug $\sin\theta$ from above,
$$E_{net}= \dfrac{2 \lambda d}{4\pi \epsilon_0 r^2} \;\hat i $$
Hence,
$$E_{net}= \dfrac{2 \lambda d}{4\pi \epsilon_0[ (d/2)^2+y^2]} \;\hat i$$
$$\boxed{E_{net}= \dfrac{8\lambda d}{4\pi \epsilon_0(d^2+4y^2)} \;\hat i}$$