The work done on an electric dipole in a uniform electric field when it is rotated from an angle $\theta_1$ to an angle $\theta_2$ is given by:

$$ W = -pE(\cos \theta_2 - \cos \theta_1) $$

where:

$ p = q \cdot d $ is the dipole moment,

$ E $ is the electric field strength,

$ \theta_1 $ and $ \theta_2 $ are the initial and final angles between the dipole moment and the electric field.

Given:

Charge $ q = 4 \, \mu \mathrm{C} = 4 \times 10^{-6} \, \mathrm{C} $

Distance $ d = 18 \, \mathrm{cm} = 0.18 \, \mathrm{m} $

Electric field $ E = 10^4 \, \mathrm{N/C} $

Initial angle $\theta_1 = 0^\circ$ (aligned with the field, equilibrium position)

Final angle $\theta_2 = 180^\circ$

First, calculate the dipole moment $ p $:

$$ p = q \cdot d = 4 \times 10^{-6} \, \mathrm{C} \cdot 0.18 \, \mathrm{m} = 7.2 \times 10^{-7} \, \mathrm{Cm} $$

Calculate the work done $ W $:

$$ W = -pE(\cos 180^\circ - \cos 0^\circ) $$

$$ W = -7.2 \times 10^{-7} \, \mathrm{Cm} \cdot 10^4 \, \mathrm{N/C} \cdot (-1 - 1) $$

$$ W = 7.2 \times 10^{-7} \, \mathrm{Cm} \cdot 10^4 \, \mathrm{N/C} \cdot 2 $$

$$ W = 1.44 \times 10^{-2} \, \mathrm{J} = 14.4 \, \mathrm{mJ} $$

Therefore, the work done on the dipole in rotating it from the equilibrium position through $180^\circ$ is 14.4 mJ (Option D).