The electric field $E$ due to a point charge is given by $E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$. Given: $q_A = 3 \times 10^{-6} \, \text{C}$, $q_B = -3 \times 10^{-6} \, \text{C}$. Distance $AB = 20 \, \text{cm} = 0.2 \, \text{m}$. The midpoint $O$ is at a distance $r = 10 \, \text{cm} = 0.1 \, \text{m}$ from both charges.

1. Field due to positive charge $q_A$ at $O$ points away from $A$ (towards $B$). Magnitude $E_A = \frac{(9 \times 10^9)(3 \times 10^{-6})}{(0.1)^2} = 2.7 \times 10^6 \, \text{N/C}$.
2. Field due to negative charge $q_B$ at $O$ points towards $B$. Magnitude $E_B = \frac{(9 \times 10^9)(3 \times 10^{-6})}{(0.1)^2} = 2.7 \times 10^6 \, \text{N/C}$. Since both fields point in the same direction (along $OB$), the net electric field is the sum of their magnitudes: $E_{net} = E_A + E_B = (2.7 + 2.7) \times 10^6 = 5.4 \times 10^6 \, \text{N/C}$, directed along $OB$.