1. Given data

• Moment of inertia about the horizontal diameter: $I = 0.8\ \text{kg m}^2$
• Magnetic moment of coil: $m = 20\ \text{A m}^2$
• Magnetic field: $B = 4\ \text{T}$
• Initial rotation from vertical position through: $\theta = 60^\circ$

2. Magnetic potential energy

For a magnetic dipole in a uniform magnetic field, $U = -mB\cos\phi$ where $\phi$ is the angle between magnetic moment $\vec m$ and magnetic field $\vec B$.

Since the coil is initially in a vertical plane, its magnetic moment is horizontal. The field is applied vertically.

So initially, $\phi_i = 90^\circ$ Hence, $U_i = -mB\cos 90^\circ = 0$

After rotating by $60^\circ$ about the horizontal diameter, the normal to the coil (and hence magnetic moment) makes angle $\phi_f = 30^\circ$ with the vertical field.

Thus, $U_f = -mB\cos 30^\circ$

3. Use conservation of energy

The decrease in magnetic potential energy becomes rotational kinetic energy: $\frac12 I\omega^2 = U_i - U_f$

So, $\frac12 I\omega^2 = 0 - (-mB\cos 30^\circ) = mB\cos 30^\circ$

Substitute values: $\frac12(0.8)\omega^2 = 20\cdot 4 \cdot \frac{\sqrt3}{2}$

$0.4\omega^2 = 40\sqrt3$

$\omega^2 = \frac{40\sqrt3}{0.4} = 100\sqrt3$

Therefore, $\omega = \sqrt{100\sqrt3} = 10\sqrt[4]{3}\ \text{rad s}^{-1}$

4. Match with options

This corresponds to: $\boxed{10\,3^{1/4}\ \text{rad s}^{-1}}$

So the correct option is C.