Three point charges of magnitude 5   μ C ,   0 . 16   μ C and 0 . 3   μ C…

Using Coulomb's law the electrostatic force on $A$ due to $C$

$F_{AC}=\frac{K{q}_A{q}_C}{{\left(r_{AC}\right)}^2}=\frac{k\times 5\times 0.3\times {10}^{-12}}{9\times {10}^{-4}}$$=\frac{9\times {10}^9\times 5\times 0.3\times {10}^{-12}}{9\times {10}^{-4}}$

$=1.5\times 10=15 \mathrm{N}$

Similarly, the force on $A$ due to $B$

$F_{AB}=\frac{K{q}_A{q}_B}{{\left(r_{AB}\right)}^2}=\frac{9\times {10}^9\times 5\times 0.16\times {10}^{-12}}{9\times {10}^{-4}}=8 \mathrm{N}$

We can see that the forces are perpendicular to each other. Therefore, the net force experienced by charge at $A$ is,

$F=\sqrt{{F_{AC}}^2+{F_{AB}}^2}$$=\sqrt{{15}^2+8^2}=\sqrt{289}=17 \mathrm{N}$