Given data

• Thickness of silver coating: $0.05\,\text{cm}$
• Area of plate: $0.05\,\text{m}^2$
• Density of silver: $7.9\,\text{g cm}^{-3}$
• Atomic mass of silver: $108\,\text{g mol}^{-1}$

Convert area into $\text{cm}^2$ Since $1\,\text{m}^2 = 10^4\,\text{cm}^2$, $0.05\,\text{m}^2 = 0.05 \times 10^4 = 500\,\text{cm}^2$

Find volume of silver deposited $\text{Volume} = \text{area} \times \text{thickness}$ $V = 500 \times 0.05 = 25\,\text{cm}^3$

Find mass of silver deposited Using $\text{mass} = \rho V$, $m = 7.9 \times 25 = 197.5\,\text{g}$

Find moles of silver $n = \frac{197.5}{108} \approx 1.83\,\text{mol}$

Find number of atoms $N = nN_A = 1.83 \times 6.022 \times 10^{23}$ $N \approx 11.0 \times 10^{23}$

Final answer The number of silver atoms deposited is $\boxed{11 \times 10^{23}}$ So the required integer is: $\boxed{11}$

Comparison with stored answer Stored correct answer = $11$

Our derived answer = $11$

Hence, the answer matches the stored correct answer.