jmc

First, we can write $\sqrt{108}=6\sqrt{3}$, $2\sqrt{12}=4\sqrt{3}$ and $2\sqrt{27}=6\sqrt{3}$. Substituting these, the expression becomes: $$\frac{4}{6\sqrt{3}+4\sqrt{3}+6\sqrt{3}}=\frac{4}{16\sqrt{3}}=\frac{1}{4\sqrt{3}}=\boxed{\frac{\sqrt{3}}{12}}$$