jmc

First, we deal with the term $\frac{2}{ab-2b^2}=\frac{2}{b(a-2b)}=\frac{4}{2b(a-2b)}.$

The quadratic $2b(a-2b),$ in $b,$ is maximized when $2b=\frac{a}{2},$ or $b=\frac{a}{4}.$ Thus, $$\frac{4}{2b(a-2b)}\ge \frac{4}{\frac{a}{2}\cdot \frac{a}{2}}=\frac{16}{a^2}.$$Then $$3a^3\sqrt{3}+\frac{2}{ab-2b^2}\ge 3a^3\sqrt{3}+\frac{16}{a^2}.$$By AM-GM, $$\begin{array}{rl}3a^3\sqrt{3}+\frac{16}{a^2}& =\frac{3a^3\sqrt{3}}{2}+\frac{3a^3\sqrt{3}}{2}+\frac{16}{3a^2}+\frac{16}{3a^2}+\frac{16}{3a^2}\\ & \ge 5\sqrt[5]{\frac{3a^3\sqrt{3}}{2}\cdot \frac{3a^3\sqrt{3}}{2}\cdot \frac{16}{3a^2}\cdot \frac{16}{3a^2}\cdot \frac{16}{3a^2}}\\ & =20.\end{array}$$Equality occurs when $\frac{3a^3\sqrt{3}}{2}=\frac{16}{3a^2}$ and $b=\frac{a}{4}.$ We can solve to get $a=\frac{2}{\sqrt{3}}$ and $b=\frac{1}{2\sqrt{3}},$ so the minimum value is $\boxed{20}.$