According to Gauss's law, the electric flux through a closed surface is related to the charge enclosed by the surface:
\[
\Phi_E = \frac{q}{\epsilon_0}
\]
where \( \Phi_E \) is the electric flux, \( q \) is the charge, and \( \epsilon_0 \) is the permittivity of free space.
Given that \( \Phi_E = -2 \times 10^4 \, \text{Nm}^2\text{C}^{-1} \) and the radius of the Gaussian surface is \( r = 8.0 \, \text{cm} \), we can solve for the charge \( q \) as:
\[
q = \Phi_E \times \epsilon_0 = (-2 \times 10^4) \times (8.85 \times 10^{-12}) = 17.7 \times 10^{-7} \, \text{C}
\]