For $\ce{HgCl2}$ ($M = \pu{271.52 g mol-1}$), the reported $K_\mathrm{sp}$ value is $10^{-15.69}$.
This means that the calculated molar solubility of $\ce{HgCl2}$ in water is $(10^{-15.69}/4)^{1/3}\cdot\pu{271.52 g mol-1} \approx \pu{0.0010 g L-1}$, or equivalently, $\pu{0.010 g cm-3}$.
However, in Pubchem, the reported molar solubility is 6.9 g/100 cc. What is the source of the large deviation between the experimental and the theoretically calculated molar solubility of $\ce{HgCl2}$?
The $\mathrm{p}K_\mathrm{sp}$ value of $15.69$ for $\ce{HgCl2}$ given by your reference must have an error. First of all, $\ce{HgCl2}$ is fairly soluble in water at $\pu{20 ^\circ C}$. Wikipedia lists it as $\pu{65.7 g L-1}$, yet lists solubility of $\ce{Hg2Cl2}$ as $\pu{3.25e-4 g L-1}$. Meantime, I found an University of Arkansas, Little Rock website, which has listed $K_\mathrm{sp}$ value of $\ce{Hg2Cl2}$ as $\pu{1.43e-18}$ ($\mathrm{p}K_\mathrm{sp} = 17.84$).
The molar mass of $\ce{Hg2Cl2}$ is $\pu{472.09 g mol-1}$. Thus, theoretically, the solubility ($s$) of $\ce{Hg2Cl2}$ can be calculated similar way you did:
$$s = \left({\frac{\pu{1.43e-18 mol3 L-3}}{4}}\right)^{\frac13}\times \pu{472.09 g mol-1} = \pu{3.35e-4 g L-1}$$
I think this is close enough. :-)
