jmc

First, we'll divide $\sqrt{6}$ into each term in the numerator of the fraction inside the big radical: $$\sqrt{\frac{\frac{4}{\sqrt{24}}+\frac{\sqrt{3456}}{9}+\sqrt{6}}{\sqrt{6}}}=\sqrt{\frac{4}{\sqrt{24}\cdot \sqrt{6}}+\frac{\sqrt{3456}}{9\cdot \sqrt{6}}+\frac{\sqrt{6}}{\sqrt{6}}}.$$Let's attack each fraction within the square root separately. First, $$\frac{4}{\sqrt{24}\cdot \sqrt{6}}=\frac{4}{\sqrt{144}}=\frac{4}{12}=\frac{1}{3}.$$The second one is trickier: $$\frac{\sqrt{3456}}{9\sqrt{6}}=\frac{\sqrt{576}}{9}=\frac{24}{9}=\frac{8}{3}.$$Finally, $\frac{\sqrt{6}}{\sqrt{6}}=1$. Adding these together, we get $$\sqrt{\frac{1}{3}+\frac{8}{3}+1}=\sqrt{\frac{1+8+3}{3}}=\sqrt{\frac{12}{3}}=\sqrt{4}=\boxed{2}.$$