Correct Answer - a
Two bowl, exerts a normal force `N` on each bead, directed along he radius line or at `60^(@)` above the horizontal. Consider the free-body diagram of the bead on the left with electric force `F_(e)` applied.
`SigmaF_(y)= N sin 60^(@)-mg-0, implies N= mg//sin 60^(@)`
`SigmaF_(X)=-F_(e )+N cos 60^(@)=0`
`implies (Kq^(2))/(R^(2))= N cos 60^(@)= (mg)/(tan 60^(@))= (mg)/(sqrt(3))`
Thus `q=R((mg)/(K sqrt(3)))^(1//2)`