At $310 \mathrm{~K}$, the solubility of $\mathrm{CaF}_{2}$ in water is $2.34 \times 10^{-3} \mathrm{~g} / 100 \mathrm{~mL}$. The solubility product of $\mathrm{CaF}_{2}$ is ____________ $\times 10^{-8}(\mathrm{~mol} / \mathrm{L})^{3}$. (Give molar mass : $\mathrm{CaF}_{2}=78 \mathrm{~g} \mathrm{~mol}^{-1}$)