How many $Ag$ atoms are there in $0.0001$ grams of $Ag$?

Molar mass of $Ag$ is $108$ grams per mole.

For $108g⟶6.023\times {10}^{23}$ atoms

For $0.00001g⟶?\left(x\right)$

$x=\frac{0.0001\times 6.023\times {10}^{23}}{108}=\frac{6.023\times {10}^{19}}{108}=5.5768\times {10}^{17}$ atoms

$\therefore 5.5768\times {10}^{17}$ atoms present in $0.0001g$ of silver.