$\sqrt[3]{2}$ solLet $BCA$ be $\theta$. Then we know that $CDB$ is $\theta-30$. Using Law of Sines on $BCD$ we get that $2=\frac{BC}{\sin{(\theta-30)}}$. Using the sine subtraction formula, and evaluating $\sin{\theta}$ and $\cos{\theta}$ from triangle $ABC$ we get that $2\sqrt{3}-2BC=AC\cdot BC$. By Gougu theorem on $ABC$ we know that $BC=\sqrt{AC^2-1}$. Thus plugging this in and solving the equation we get that $AC=\sqrt[3]{2}$ as the only real solution.