Answer
$1.4\times 10^8 N/C$.
Work Step by Step
The electric field due to the positive charge (#1) points away from it. The electric field due to the negative charge (#2) points toward it. Because both fields point in the same direction, the magnitudes are added.
$$E=E_1+E_2$$
$$E=\frac{k|Q_1|}{r_1^2}+\frac{k|Q_2|}{r_2^2}$$
$$=(8.99\times10^9\frac{N\cdot m^2}{C^2})(\frac{8.0\times10^{-6}C}{(0.030m)^2}+\frac{5.8\times10^{-6}C}{(0.030m)^2})$$
$$=1.4\times 10^8 N/C$$
The direction is toward the negative charge.