$100\,g$ of chlorohydrocarbon has $3.55\,g$ of chlorine.

$1\,g$ of chlorohydrocarbon will have $\frac{{3.55}}{{100}} = 0.0355\,g$ of chlorine.

Atomic wt. of $Cl = 35.5\,g/mol$

Number of moles of $Cl$ $ = \frac{{0.0355\,g}}{{35.5\,g/mol}} = 0.001$ $mole$

Number of atoms of $Cl = 0.001\,mole$ $\times 6.023 \times 10^{23}\,mol^{-1}$

$= 6.023 \times 10^{20}$