The correct order of solubility of the given salts in water at 298 K is SaltKsp at 298 K AgBr$5.0 \times 10^{-13}$ Zn(OH)2$1.0 \times 10^{-15}$ Hg2Cl2$1.3 \times 10^{-18}$

Since all three salts are of the $AB_2$ or similar types (calculating approximate molar solubility $s$ from $K_{sp}$), higher $K_{sp}$ directly corresponds to higher solubility for salts having comparable stoichiometry formulas, but let's check: AgBr is $AB$ type ($s = \sqrt{K_{sp}} \approx 7.0 \times 10^{-7}$), $\text{Zn(OH)}_2$ is $AB_2$ type ($s = (K_{sp}/4)^{1/3} \approx 6.3 \times 10^{-6}$), $\text{Hg}_2\text{Cl}_2$ is $A_2B_2$ type ($s = (K_{sp}/4)^{1/3}$). Comparing calculated solubilities gives AgBr > $\text{Zn(OH)}_2$ > $\text{Hg}_2\text{Cl}_2$.