A closed surface encloses n electric dipoles. The net electric flux...

An electric dipole consists of two equal and opposite charges, $+q$ and $-q$.

The net charge of a single electric dipole is $+q - q = 0$.

For $n$ electric dipoles enclosed in a closed surface, the total enclosed charge is $Q_{enclosed} = n \times 0 = 0$.

According to Gauss's law, the net electric flux through a closed surface is given by $\Phi = \dfrac{Q_{enclosed}}{\epsilon_0}$.

Substituting $Q_{enclosed} = 0$, we get $\Phi = 0$.

Answer: $0$