1. The problem asks to write $9^{-2}$ as a fraction without indices. 2. Recall the rule for negative exponents: $a^{-n} = \frac{1}{a^n}$. 3. Applying this rule, we get: $$9^{-2} = \frac{1}{9^2}$$ 4. Calculate $9^2$: $$9^2 = 9 \times 9 = 81$$ 5. Substitute back: $$9^{-2} = \frac{1}{81}$$ 6. Therefore, the number that should go in the box is $81$.