Answer
$E_{net} = 160~N/C$
Work Step by Step
By symmetry, the electric fields due to particle 1 and particle 2 cancel out.
The net electric field is equal to the electric field due to particle 3.
We can find the magnitude of $E_{net}$:
$E_{net} = \frac{1}{4~\pi~\epsilon_0}~\frac{\vert q \vert}{r^2}$
$E_{net} = \frac{1}{4~\pi~\epsilon_0}~\frac{\vert q \vert}{(a~cos~45^{\circ})^2}$
$E_{net} = \frac{1}{(4~\pi)~(8.854\times 10^{-12}~F/m)}~\frac{(2)(1.60\times 10^{-19}~C)}{[(6.00\times 10^{-6}~m)~cos~45^{\circ}]^2}$
$E_{net} = 160~N/C$
