If there was no hole, charge Q and radius r of the sphere
Charge Density (conducting sphere) [math](\sigma)=\dfrac{Q}{4 \pi r^2}[/math]
Electric Field=[math]E=\dfrac{Q}{4 \pi \epsilon_0 r^2}[/math] in the radial direction
Then instead of cutting a hole, you can put a surface charge of
[math]-\sigma[/math] there.
The field is then =[math]\dfrac {\sigma}{\epsilon_0}[/math] plus the field due to the additional surface charge density of [math]-\sigma[/math]. That additional field is
[math]-\dfrac{\sigma}{2 \epsilon_0}[/math],
so the total field is= [math]\dfrac{\sigma}{2 \epsilon_0} =\dfrac{Q}{8 \pi \epsilon_0 r^2}[/math] in the radial direction.