Correct Answer - `(q^2)/(8 pi epsilon_0) ((1)/a - (1)/b)`
Potential at center O due to induced charges is
`V_1 = - q/(4 pi epsilon_0 a) + q/(4 pi epsilon_0 b)`
`V_(oo)=0`
small work done to transfer a small charge dq from center to infinity is
`d W = d q (V_oo - V_1) = (1)/(4 pi epsilon_0) [(1)/a - (1)/b] q d q`
or `W = int d W = (q^2)/(8 pi epsilon_0) [(1)/a - (1)/b]`.