The answer does involve a reciprocal, but not $\displaystyle\frac{{1}}{{6}}\usepackage{cmbright}\Huge\$

Since $\displaystyle{{n}}^{{-{x}}}=\frac{{1}}{{{n}}^{{x}}}\usepackage{cmbright}\Huge\$ , this tells you that $\displaystyle{{3}}^{{-{2}}}=\frac{{1}}{{{3}}^{{2}}}\usepackage{cmbright}\Huge\$