To solve this problem, we need to use Faraday's laws of electrolysis and the concept of molar volume at STP.$\\$Given:$\\$• Atomic weight of silver (Ag) $= 108 \\$• Volume of $O_2$ displaced at STP $= 5600$ mL$\\$At STP, 1 mole of any gas occupies 22400 mL. Therefore, the moles of $O_2$ displaced are:$\\$Moles of $O_2 = \frac{5600}{22400} = \frac{1}{4} \\$The reaction for the formation of $O_2$ is:$\\ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \\$This shows that 1 mole of $O_2$ is produced by 4 moles of electrons.$\\$Therefore, $\frac{1}{4}$ mole of $O_2$ is produced by $\frac{1}{4} \times 4 = 1$ mole of electrons.$\\$According to Faraday's laws, 1 mole of electrons (1 Faraday) will deposit 1 mole of silver.$\\$The weight of 1 mole of silver is its atomic weight, which is 108 g.$\\$Therefore, the weight of silver displaced is $108$ g.$\\$Thus, the correct option is 4.