Calculate the resultant force of the charge as seen due to the charges A and B.

Since $Q_A$ is positive and $Q_B$ is negative, the force is attractive, meaning $F_{BA}$ acts to the left (towards A).

$$F_{BA} = k \frac{|Q_A Q_B|}{r_{AB}^2}$$ $$F_{BA} = (9 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{|(+8 \times 10^{-6} \text{ C})(-10 \times 10^{-6} \text{ C})|}{(0.12 \text{ m})^2}$$ $$F_{BA} = (9 \times 10^9) \frac{80 \times 10^{-12}}{0.0144}$$ $$F_{BA} = \frac{720 \times 10^{-3}}{0.0144}$$ $$F_{BA} = 50 \text{ N}$$

The direction of $F_{BA}$ is to the left.

Since $Q_B$ is negative and $Q_C$ is positive, the force is attractive, meaning $F_{BC}$ acts to the right (towards C).

$$F_{BC} = k \frac{|Q_B Q_C|}{r_{BC}^2}$$ $$F_{BC} = (9 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{|(-10 \times 10^{-6} \text{ C})(+12 \times 10^{-6} \text{ C})|}{(0.06 \text{ m})^2}$$ $$F_{BC} = (9 \times 10^9) \frac{120 \times 10^{-12}}{0.0036}$$ $$F_{BC} = \frac{1080 \times 10^{-3}}{0.0036}$$ $$F_{BC} = 300 \text{ N}$$

The direction of $F_{BC}$ is to the right.

Let's consider the direction to the right as positive.
The resultant force $F_{net}$ is the vector sum of $F_{BC}$ and $F_{BA}$.

$$F_{net} = F_{BC} - F_{BA}$$ $$F_{net} = 300 \text{ N} - 50 \text{ N}$$ $$F_{net} = 250 \text{ N}$$

Since the result is positive, the resultant force is directed to the right (towards charge C).

The resultant force on charge B is $\text{250 N to the right}$.