[tex]\frac{\left(0,125\right)^2+2^{-3}}{\left(\frac{2}{3}\right)^{-2}}=\\\\\\\frac{\left ( \frac{125^{\div125}}{1000^{\div125}} \right )^2+\left ( \frac{1}{2} \right )^3}{\left ( \frac{3}{2} \right )^2}=\\\\\\\frac{\left ( \frac{1}{8} \right )^2+\left ( \frac{1}{8} \right )}{\left ( \frac{9}{4} \right )}=\\\\\\\frac{\frac{1}{64}+\frac{1}{8}}{\frac{9}{4}}=\\\\\\\frac{9}{64}\div\frac{9}{4}=\\\\\\\frac{9}{64}\times \frac{4}{9}=\\\\\\\frac{4}{64}=\\\\\\\boxed{\frac{1}{16}}[/tex]
Portanto, alternativa b.