Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero.
Let E is the electric field just outside the conductor, q is the electric charge, a is the charge density and ${\mathrm{\epsilon }}_0$ is the permittivity of free space.
Charge q = $\mathrm{\sigma }$ X ds

According to Gauss's law, flux, $\mathrm{\varnothing }=E\cdot ds=\frac{q}{{ϵ}_0}$

$\begin{array}{l}\Rightarrow E.ds=\frac{\sigma \times ds}{{ϵ}_0}\\ \therefore E=\frac{\sigma }{2{ϵ}_0}\hat{n}\end{array}$

Therefore, the electric field just outside the conductor is $\frac{\sigma }{2{ϵ}_0}\hat{n}$. This field is a superposition of field due to the cavity E and the field due to the rest of the charged conductor E. These fields are equal and opposite inside the conductor and equal in magnitude and direction outside the conductor.  $\therefore E+E^{\mathrm{\prime }}=E$

$\Rightarrow E^{\mathrm{\prime }}=\frac{E}{2}=\frac{\sigma }{2{ϵ}_0}\hat{n}$

Hence, the field due to the rest of the conductor is $\frac{\sigma }{{ϵ}_0}\hat{n}$.