Correct Answer - b
Electric field at point `P` due to charge of ring is
`E=(kQx)/((R^(2)+x^(2))^(3//2))`
At `x= R:E= (kQ)/(2sqrt(2)R^(2))` directed towards the centre.
Electric field at `P` due to centre charge: `(kq)/(R^(2))` for net field to be zero at `P`:
`(kq)/(R^(2))=(kQ)/(2sqrt(2)R^(2))implies q=(Q)/(2sqrt(2))=(Q)/(4)sqrt(2)`