Question

What is the net electric flux through the torus (i.e., doughnut shape) of the figure (Figure 1) ? Assume that Q=300nC and q=−8.0nC

Answer 1

Since Q is outside, it wont add to net flux. So ignore it.
As of q, it is inside the torus.
So From Gauss' Law, net flux = integration(E.dS) = q/epsilon_0
Since we dont know Eletric field, we can find flux simply as q/epsilon_0, as q is the charge enclosed in the given torus.
Net flux =−8.0nC/epsilon_0 =-8.0e-9 C/8.85418782 × 10^-12 m^-3 kg^-1 s⁴ A² = -903.53 Vm

Answer 2

If you have -2q it is -225.98