Answer
$1.15\times 10^6 N/C$, toward the $-38.6 \mu C$ charge.
Work Step by Step
The two $-27.0 \mu C$ negative charges opposite from each other will create fields at the center that cancel each other. Only the two remaining charges matter.
The electric field due to the $-38.6 \mu C$ charge (#1) points toward it. The electric field due to the $-27.0 \mu C$charge (#2) points toward it. Because these two fields point in opposite directions, the magnitudes should be subtracted. The distance from each charge to the center is $(42.5 cm/\sqrt{2})$.
$$E=E_1-E_2$$
$$E=\frac{k|Q_1|}{r_1^2}-\frac{k|Q_2|}{r_2^2}$$
$$=(8.99\times10^9\frac{N\cdot m^2}{C^2})(\frac{38.6\times10^{-6}C}{(0.425 m/\sqrt{2})^2}-\frac{27.0\times10^{-6}C}{(0.425 m/\sqrt{2})^2})$$
$$=1.15\times 10^6 N/C$$
The direction is toward the $-38.6 \mu C$ charge.
