The solubility of $\mathrm{CdSO}_4$ in water is $8.0 \times 10^{-4} \mathrm{~mol} \mathrm{L}^{-1}$. Its solubility in $0.01 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$ solution is $\_\_\_\_$ $\times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}$. (Round off to the Nearest integer) (Assume that solubility is much less than $0.01 \mathrm{M})$