The electric potential $V$ due to a short electric dipole at a distance $r$ from the centre and at an angle $\theta$ with the dipole axis is given by the formula: $V = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2}$ [NCERT Class 12, Physics Part I, Sec 2.4, Eq 2.15]

1. Given Data: Dipole moment, $p = 16 \times 10^{-9} \text{ C m}$ Distance, $r = 0.6 \text{ m}$ Angle, $\theta = 60^{\circ}$ Coulomb constant, $k = 9 \times 10^9 \text{ N m}^2\text{C}^{-2}$

2. Calculation: Substitute the values into the formula: $V = \frac{(9 \times 10^9) \times (16 \times 10^{-9}) \times \cos(60^{\circ})}{(0.6)^2}$ Simplify the expression (note $10^9 \times 10^{-9} = 1$ and $\cos 60^{\circ} = 0.5$): $V = \frac{9 \times 16 \times 0.5}{0.36}$ $V = \frac{144 \times 0.5}{0.36} = \frac{72}{0.36}$ $V = \frac{7200}{36} = 200 \text{ V}$