Answer
$\frac{27}{4}$
Work Step by Step
According to the negative exponent rule, $a^{-n}=\frac{1}{a^{n}}$.
Therefore, $\frac{2^{-2}}{3^{-3}}=\frac{1}{\frac{2^{2}}{3^{3}}}=1\div\frac{2^{2}}{3^{3}}=1\times\frac{3^{3}}{2^{2}}=1\times\frac{27}{4}=\frac{27}{4}$.