Answer
$-16.97\mu C$
Work Step by Step
Let $q$ charge must be placed at $x=24\;m$.
The magnitude of net electric field at origin due to charges at $x=8\;m$ and $x=16\;m$ is given by
$E=\frac{1}{4\pi\epsilon_0}[\frac{(6\times10^{-6})^2}{8^2}-\frac{(4\times10^{-6})^2}{16^2}]\;N/C=\frac{1}{4\pi\epsilon_0}\frac{10^{-12}}{2}\;N/C$, directed towards negative x-axis.
Therefore, the charge at $x=24\;m$ should be negative so that the electric field due to it is directed towards positive x-axis. The electric field at origin due to the charge at $x=24\;m$ is given by
$E^{'}=\frac{1}{4\pi\epsilon_0}\frac{q^2}{24^2}\;N/C$
Any charge placed at the origin would experience no electrostatic force. This implies net electric field at origin is zero. Therefore,
$\frac{1}{4\pi\epsilon_0}[\frac{q^2}{24^2}-\frac{10^{-12}}{2}]=0$
or, $\frac{q}{24}=\frac{10^{-6}}{\sqrt 2}$
or, $q=16.97\times 10^{-6}\;C=16.97\mu C$
Taking the sign convention, the required charge is $-16.97\mu C$
