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