Answer
0.16 g Ag
Work Step by Step
t= 9300 s
$Q= It=0.015\,A\times9300\,s=139.5\,C$
According to the reaction:
$Ag^{+}+e^{-}\rightarrow Ag$
We require 96487 C to deposit 1 mol or 107.87 g of Ag.
Therefore, mass of Ag deposited=
$\frac{107.87\,g\,mol^{-1}\times139.5\,C}{96487\,C\,mol^{-1}}= 0.16\,g$
