35 ppm CO,means $10^6g$ of air contains =$35g\;CO$
$\therefore$ Volume of air =$\large\frac{\text{mass of air}}{\text{density of air}}$
$\Rightarrow \large\frac{10^6}{1.3}$$=7.69\times 10^5L$
Now since $7.69\times 10^5L$ of air has =35g CO
$\therefore$ 1L of air has =$35/7.69\times 10^5\;CO$
$\Rightarrow 4.55\times 10^{-5}g\;CO$