Volume of $Ag$ deposited $= (0.5 \times 100 \times 100)cm^2 \times 0.005\;cm=25\;cm$

Mass of $Ag$ deposited =Volume $\times$ density

$\qquad=25 \times 7.9=197.5$

$108\;gm$ of $Ag$ contain $=\large\frac{6.023 \times 10^{23}}{108} \times$$ 197$

$\qquad= 10.9 \times 10^{23}=1.1 \times 10^{24}$

Hence c is the correct answer.