The time period to coat a metal surface of $80{\mathrm{cm}}^2$ with $5\times {10}^{-3}\mathrm{cm}$ thick layer of silver (density 1.05 g cm^-3) with the passage of 3A current through a silver nitrate solution is

Volume = area  thickness 
$$\begin{gathered} 80{\mathrm{cm}}^2\times 5\times {10}^{-3}\mathrm{cm} \\ =0.4{\mathrm{cm}}^3 \end{gathered}$$
Given $\mathrm{d}=1.05\mathrm{g}/{\mathrm{cm}}^3$
$$\begin{gathered} \Rightarrow 1.05\mathrm{g}\to 1{\mathrm{cm}}^3 \\ ?\to 0.4{\mathrm{cm}}^3 \\ \mathrm{x}=\frac{1.05\times 0.4}{1}=0.42\mathrm{g} \end{gathered}$$
For Ag metal  1F (or)  $96500\mathrm{C}\to 108\mathrm{g}$ of Ag
$?\to 0.42\mathrm{g}$Ag
$\mathrm{y}=\frac{96500\times 0.42}{108}=375.27\mathrm{C}$
We know that $\mathrm{q}=\mathrm{C}\times \mathrm{t}$
$$\begin{gathered} 375.27=3\times \mathrm{t} \\ \mathrm{t}=\frac{375.27}{3}=125\sec \end{gathered}$$