\[\frac{\left( 2^{- 2} \right)^{4} \cdot 16^{2}}{2^{3}} = \frac{2^{- 8} \cdot \left( 2^{4} \right)^{2}}{2^{3}} =\]
\[= \frac{2^{- 8} \cdot 2^{8}}{2^{3}} = \frac{1}{8}.\]