Obliczmy wartość tej liczby:   

$\frac{2^{-7}\cdot {16}^3}{{\left(8^3\cdot 2^2\right)}^4}\cdot {\left(\frac{1}{4}\right)}^{-2}=\frac{2^{-7}\cdot {\left(2^4\right)}^3}{{\left({\left(2^3\right)}^3\cdot 2^2\right)}^4}\cdot 4^2=\frac{2^{-7}\cdot 2^{12}}{{\left(2^9\cdot 2^2\right)}^4}\cdot 2^4=$  

$=\frac{2^5}{{\left(2^{11}\right)}^4}\cdot 2^4=\frac{2^9}{2^{44}}=2^{-35}$  


Odp.: B