http://www.smartphysics.com/images/content/EM/02/h2_arcA.png

First from symmetry considerations, you have that the electric field is directed along the line y=x. So you need only to compute the total Electric field in the origin: this is

From the figure the charge Q is uniformly distributed on the cyan quarter circle, thus the linear density is 2Q/\pi a. Thus the electric field is
\int_0^{\pi/2} \frac{2Q/\pi}{a^2} d\theta=Q/a^2
and the x component is \frac{Q}{\sqrt{2}a^2} .