Solve this :

$$\begin{gathered} \mathrm{Q}. \mathrm{The} \mathrm{solubility} \mathrm{product} \mathrm{of} \mathrm{AgBr} \mathrm{is} 4.9\times {10}^{-9}. \mathrm{The} \mathrm{solubility} \mathrm{of} \mathrm{AgBr} \mathrm{will} \mathrm{be} : \\ \left(1\right) 7\times {10}^{-4} \mathrm{mole}/\mathrm{litre} \\ \left(2\right) 7\times {10}^{-5} \mathrm{g}/\mathrm{litre} \\ \left(3\right) 1.316\times {10}^{-2 }\mathrm{g}/\mathrm{litre} \\ \left(4\right) 1\times {10}^{-3} \mathrm{mole}/\mathrm{litre} \end{gathered}$$