Explanation

(d)
$Number\ of\ oxygen\ atom\ in\ 2\ g\ of\ CO$
$\underset{\ \ \ \ \ \ }{= \frac{2}{28}} \times 6.022 \times 10^{23} \times 1$
$Number\ of\ oxygen\ atom\ in\ 2\ g\ of\ {CO}_{2}$
$= \frac{2}{44} \times 6.022 \times 10^{23} \times 2$
$Number\ of\ oxygen\ atom\ in\ 2\ g\ of\ {SO}_{2}$
$= \frac{2}{64}\ \times 6.022 \times 10^{23} \times 2$
Number of oxygen atom in 2 g of $H_{2}O$
$= \frac{2}{18} \times 6.022 \times 10^{23} \times 1$
Hence, 2 g of $H_{2}O$ has maximum number of atoms of oxygen.