Correct Answer - C
Force one sphere `C` due to `A, F_(AC)` = Force on sphere
`C` due to `B, F_(BC) =(1)/(4pi in_(0)) (q^(2))/((2R)^(2))`
Now `F_(AC)` and `F_(BC)` are inclined at `60^(@)` to each other, therefore resultant force on `C` is
`F_(R) = sqrt(F_(AC)^(2) + F_(BC)^(2) + 2 F_(AC) F_(BC) cos 60^(@))`
`= sqrt(F^(2) + F^(2) + 2F^(2) cos 60^(@))`
`= sqrt(3) F` [Taking `F_(AC) = F_(BC) = F`]
`= (1)/(4pi in_(0)) (sqrt(3))/(4) ((q)/(R))^(2)`