Three point charges q, - 2q and 2q are placed on x axis at a...

The net force on the given charge will be the resultant of the forces due to the other two charges.

Hence, the net force $\left({F}_{n}\right)$ can be calculated as follows:

$\begin{cases}{F}_{n} & = & \dfrac{1}{4\pi{ε}_{0}}\dfrac{4{q}^{2}}{{\left(\dfrac{R}{4}\right)}^{2}} - \dfrac{1}{4\pi{ε}_{0}}\dfrac{2{q}^{2}}{{\left(\dfrac{3R}{4}\right)}^{2}} \\ & = & 9 \times {10}^{9} \times \dfrac{16{q}^{2}}{{R}^{2}}\left[4 - \dfrac{2}{9}\right] \\ & = & 9 \times {10}^{9} \times \dfrac{16 \times {\left(2 \times {10}^{- 6}C\right)}^{2}}{{\left(0 . 02m\right)}^{2}} \times \dfrac{34}{9} \\ & = & 5440N\end{cases}$