$$2^{-\frac{1}{3}} \cdot 27^{-\frac{1}{3}} \cdot \sqrt[3]{16} = 2^{-\frac{1}{3}} \cdot (3^3)^{-\frac{1}{3}} \cdot \sqrt[3]{2^4} = 2^{-\frac{1}{3}} \cdot 3^{-1} \cdot 2^{\frac{4}{3}} = \frac{2^{-\frac{1}{3}} \cdot 2^{\frac{4}{3}}}{3} = \frac{2^{-\frac{1}{3} + \frac{4}{3}}}{3} = \frac{2^{\frac{3}{3}}}{3} = \frac{2}{3}$$
Ответ: $$\frac{2}{3}$$