We know that net electric flux can be given by
ϕnet=→E.→S
Since, the electric field is radially outwards, the angle between electric field and area vector will be 0∘
⟹ϕnet=E.S.cos(0∘)
⟹ϕnet=αR4πR2
⟹ϕnet=α4πR3-----(i)
By Gauss law, we know that
ϕnet=Qinϵ0
By using (i), we can write
4παR3=Qinϵ0
⟹Qin=4πϵ0αR3
⟹ Qin=19×109×100×(0.3)3
∴Qin=3×10−10 C
⇒m=3