Answer
16
Work Step by Step
We know that $a^{\frac{m}{n}}=(\sqrt[n] a)^{m}$ (where m and n are positive integers greater than 1 with $\frac{m}{n}$ in simplest form and $\sqrt[n] a$ is a real number).
Therefore, $(-64)^{\frac{2}{3}}=(\sqrt[3] -64)^{2}=(-4)^{2}=16$
We know that $\sqrt[3] -64=-4$ because $(-4)^{3}=-64$