Answer:
B) 1 over (6 to the power of 9)
Step-by-step explanation:
[tex]\left(6^{-2}\cdot6^5\right)^{-3}\qquad\text{use}\ a^n\cdot a^n=a^{n+m}\\\\=(6^{-2+5})^{-3}=(6^3)^{-3}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=6^{(3)(-3)}=6^{-9}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\\\\=\dfrac{1}{6^9}[/tex]