To solve this problem, we need to calculate the electric potential due to a dipole at a given point.$\\$The formula for the electric potential $V$ due to a dipole at a point $P$ is given by:$\\ V = \frac{1}{4\pi\epsilon_0} \cdot \frac{p \cdot \cos\theta}{r^2} \\$where:$\\$• $p = 16 \times 10^{-9}$ C m (dipole moment)$\\$• $\theta = 60^{\circ}$ (angle with the dipole axis)$\\$• $r = 0.6$ m (distance from the center of the dipole)$\\$• $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9$ N m$^2$ C$^{-2} \\$Substitute the known values into the formula:$\\ V = 9 \times 10^9 \cdot \frac{16 \times 10^{-9} \cdot \cos 60^{\circ}}{(0.6)^2} \\$Calculate $\cos 60^{\circ}: \\ \cos 60^{\circ} = \frac{1}{2} \\$Substitute $\cos 60^{\circ}$ into the equation:$\\ V = 9 \times 10^9 \cdot \frac{16 \times 10^{-9} \cdot \frac{1}{2}}{0.36} \\$Simplify the expression:$\\ V = 9 \times 10^9 \cdot \frac{8 \times 10^{-9}}{0.36} \\ V = \frac{72 \times 10^0}{0.36} \\ V = 200$ V$\\$Therefore, the electric potential at the given point is $200$ V. This corresponds to Option 1.