jmc

From right triangle $ABD$, we have $\sin A=\frac{BD}{AB}=\frac{BD}{24}$. Since $\sin A=\frac{2}{3}$, we have $\frac{2}{3}=\frac{BD}{24}$, so $BD=\frac{2}{3}\cdot 24=16$.

From right triangle $BCD$, we have $\sin C=\frac{BD}{BC}=\frac{16}{BC}$. Since $\sin C=\frac{1}{3}$, we have $\frac{16}{BC}=\frac{1}{3}$. Therefore, we have $BC=3\cdot 16=48$. Finally, the Pythagorean Theorem gives us $$\begin{array}{rl}CD& =\sqrt{B{C}^2-B{D}^2}\\ & =\sqrt{48^2-16^2}\\ & =\sqrt{(3\cdot 16{)}^2-16^2}\\ & =\sqrt{9\cdot 16^2-16^2}=\sqrt{8\cdot 16^2}=2\cdot 16\sqrt{2}=\boxed{32\sqrt{2}}.\end{array}$$