#### Answer

The strength of the electric field is $7640~N/C$ and it is directed to the left.

#### Work Step by Step

By symmetry, the vertical components of the electric field from each charge cancel out. The net electric field is the sum of the horizontal components of the electric field from each charge, which are directed toward the left.
$E = 2~\frac{k~q}{r^2}$
$E = 2~\frac{(9.0\times 10^9~N~m^2/C^2)(3.0\times 10^{-9}~C)}{[(0.050~m)(\sqrt{2})]^2}~cos(45^{\circ})$
$E = 7640~N/C$
The strength of the electric field is $7640~N/C$ and it is directed to the left.