What will be the strength of the electric field if the Gaussian surface does not include any net charge?

There is no way to know. All we can say is that if the Gaussian surface does not include (enclose) any net charge, the flux of the electric field through this surface is zero. This follows from the Gauss Law:

`Phi_E = Q/(epsilon_0)` . Here, `Phi_E` is the electric flux through a closed surface, Q is the net charge enclosed by the surface, and `epsilon_0` is dielectric permittivity of vacuum. If the enclosed net charge is zero, then the flux ix zero.

However, this does not necessarily imply that the electric field is also zero. The electric flux is the surface integral:

`Phi_E = int_S vecE*dvecS`

which can be zero even if the electric field is not zero. The value of this integral depends on how the electric field lines are located in relationship to the surface.

So, it is impossible to determine the strength of the electric field simply from the fact that the net charge enclosed by the specific Gaussian surface is zero.