The work done $W$ in rotating an electric dipole in a uniform electric field is given by:

$W = pE(1 - \cos\theta)$,

where $p$ is the dipole moment, $E$ is the strength of the electric field, and $\theta$ is the angle the dipole is rotated through.

In this case, the dipole moment $p$ is $6.0 \times 10^{-6} ~\mathrm{C m}$, the electric field $E$ is $1.5 \times 10^{3} ~\mathrm{NC}^{-1}$, and the angle $\theta$ is $180^{\circ}$.

Substituting these values into the formula gives:

$W = 6.0 \times 10^{-6} ~\mathrm{C m} \times 1.5 \times 10^{3} ~\mathrm{NC}^{-1} \times (1 - \cos180^{\circ})$.

Since $\cos180^{\circ} = -1$, the equation becomes:

$W = 6.0 \times 10^{-6} ~\mathrm{C m} \times 1.5 \times 10^{3} ~\mathrm{NC}^{-1} \times (1 - (-1))$,

$W = 6.0 \times 10^{-6} ~\mathrm{C m} \times 1.5 \times 10^{3} ~\mathrm{NC}^{-1} \times 2$,

$W = 18.0 \times 10^{-3} ~\mathrm{J} = 18.0 ~\mathrm{mJ}$.

So the work done in rotating the dipole by $180^{\circ}$ in this field is 18.0 millijoules.