The solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10–12 and 5.0 × 10–13 respectively.

Let $s$ be the solubility of $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$.

Then, $\mathrm{Ag}_{2} \mathrm{CrO}_{4} \longleftrightarrow \mathrm{Ag}^{2+}+2 \mathrm{CrO}_{4}^{-}$

$K_{s p}=(2 s)^{2} \cdot s=4 s^{3}$

$1.1 \times 10^{-12}=4 s^{3}$

$s=6.5 \times 10^{-5} \mathrm{M}$

Let $s^{\prime}$ be the solubility of $\mathrm{AgBr}$.

$\mathrm{AgBr}_{(s)} \longleftrightarrow \mathrm{Ag}^{+}+\mathrm{Br}^{-}$

$K_{s p}=s^{\prime 2}=5.0 \times 10^{-13}$

$\therefore s^{\prime}=7.07 \times 10^{-7} \mathrm{M}$

Therefore, the ratio of the molarities of their saturated solution is $\frac{s}{s^{\prime}}=\frac{6.5 \times 10^{-5} \mathrm{M}}{7.07 \times 10^{-7} \mathrm{M}}=91.9$.