$$a^{16} \cdot (a^9)^{-2} = a^{16} \cdot a^{-18} = a^{16-18} = a^{-2} = \frac{1}{a^2}$$
При $$a = \frac{1}{8}$$:
$$\frac{1}{a^2} = \frac{1}{(\frac{1}{8})^2} = \frac{1}{\frac{1}{64}} = 64$$
Ответ: 64