To determine the potential energy of a magnetic dipole in a uniform magnetic field, we start with the torque equation:

$ \tau = M \times B = MB \sin 60^\circ $

Given that $\tau = 80 \sqrt{3} \, \text{N m}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$, we can equate and solve for $MB$:

$ MB \frac{\sqrt{3}}{2} = 80 \sqrt{3} $

Solving for $MB$, we find:

$ MB = 160 $

The potential energy $U$ of the dipole in the magnetic field is given by:

$ U = -M \cdot B = -MB \cos 60^\circ $

Substituting in $\cos 60^\circ = \frac{1}{2}$, we calculate:

$ U = -160 \times \frac{1}{2} = -80 \, \text{J} $

Thus, the potential energy of the dipole is $-80 \, \text{J}$.