Solution:
$54 \sqrt{3}$

We can first compare $54 \sqrt{3}$ and $144$. Note that $\sqrt{3} < 2$ and $\frac{144}{54} = \frac{8}{3} > 2$. Hence, $54 \sqrt{3}$ is less.

Now, we wish to compare this to $108 \sqrt{6} - 108 \sqrt{2}$. This is equivalent to comparing $\sqrt{3}$ to $2(\sqrt{6} - \sqrt{2})$. We claim that $\sqrt{3} < 2(\sqrt{6} - \sqrt{2})$.

To prove this, square both sides to get $3 < 4(8 - 4 \sqrt{3})$ or $\sqrt{3} < \frac{29}{16}$ which is true because $\frac{29^{2}}{16^{2}} = \frac{841}{256} > 3$.

We can reverse this sequence of squarings because, at each step, we make sure that both our values are positive after taking the square root. Hence, $54 \sqrt{3}$ is the smallest.