Correct answer is (a) \(\frac {q}{r}\)
Easy explanation: Force on a unit point charge kept at a distance r from the charge=\(\frac {q}{r^2}\). Therefore, work done to bring that point charge through a small distance dr=\(\frac {q}{r^2}\)*(-dr). Therefore, the potential of that point is =\(\int_\alpha^r\frac {-q}{r^2}dr = \frac{q}{r}\).