Answer
$q=-7.97\mu C$
Work Step by Step
We know that:
$\phi=\frac{q}{\epsilon_{\circ}}$
This can be rearranged as:
$q=\epsilon_{\circ}\phi$
The flux $\phi$ of the particle is given by $\phi=-1.8\phi_s=-1.8(5.0\times 10^5)=-9\times 10^5\frac{N.m^2}{C}$
We plug in the known values to obtain:
$q=8.85\times 10^{-12}\times (-9\times 10^5)=-7.97\times 10^{-6}=-7.97\mu C$
