Consider an infinitesimal section of the rod of length $dx$, a distance $x$  from the left end, as shown in the figure below.

It contains charge $dq=\lambda dx$ and is a distance $r$ from $L$.

Magnitude of the field produces by $dx$ at $L$ is given by $dE=\frac{k\lambda dx}{r^2}$.

The $x$ and the $y$ components of electric field are $d{E}_{\mathrm{x}}=\frac{k\lambda dx}{r^2}\sin \theta$ and $d{E}_{\mathrm{y}}=\frac{k\lambda dx}{r^2}\cos \theta$.

Using $\theta$ as variable of integration and substituting $r=\frac{L}{\cos \theta }$ and $x=L\tan \theta$.

$\Rightarrow dx=\frac{L}{{\cos }^2\theta }d\theta$

The limits of integration are $0$ and $\frac{\pi }{r}$ radian. 

$E_{\mathrm{x}}=\frac{k\lambda }{L}{\int }_0^{\frac{\pi }{2}}\sin \theta d\theta =-\frac{k\lambda }{L}{\left[\cos \theta \right]}_0^{\frac{\pi }{2}}=\frac{k\lambda }{L}$

$E_{\mathrm{y}}=\frac{k\lambda }{L}{\int }_0^{\frac{\pi }{2}}\cos \theta d\theta =\frac{k\lambda }{L}{\left[\sin \theta \right]}_0^{\frac{\pi }{2}}=\frac{k\lambda }{L}$

Now, electric field is $E=\sqrt{E_x^2+E_y^2}=\sqrt{{\left(\frac{k\lambda }{L}\right)}^2+{\left(\frac{k\lambda }{L}\right)}^2}$

$=\frac{\sqrt{2}k\lambda }{L}$

$=\frac{\sqrt{2}\lambda }{4\pi {\epsilon }_{0}L}=\frac{\lambda }{2\sqrt{2}\pi {\epsilon }_{0}L}$