$(a)\;6\times10^{19}\qquad(b)\;6\times10^{18}\qquad(c)\;6\times10^{17}\qquad(d)\;6\times10^{13}$

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At constant V and T

$\large\frac{P_1}{n_1} = \large\frac{P_2}{n_2}$

$\large\frac{100}{\large\frac{12}{120}} = \large\frac{0.01}{n_2}$

$(n_1 = \large\frac{12}{120})$

$n_2 = \large\frac{0.01}{1000}$

$\;\;\;\;=1\times10^{-5}$

No. of molecules left = $6.02\times10^{23}\times10^{-5}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=6.02\times10^{18}$

Hence answer is (b)

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