The relation between number of atoms in the bulb X and Y is

$\text{16 g of O = 1 mol of O =\hspace{0.17em}6.023}\times {\text{10}}^{\text{23}}\text{\hspace{0.17em}oxygen\hspace{0.17em}atoms}$

${\text{48 g of O}}_{\text{3}}{\text{ = 1 mol of O}}_{\text{3}}=3\text{\hspace{0.17em}mol\hspace{0.17em}of oxygen atoms}$

${\text{16 g of O}}_{\text{3}}{\text{ = 1/3\hspace{0.17em}mol of O}}_{\text{3}}=1\text{\hspace{0.17em}mol\hspace{0.17em}of "O atoms" =\hspace{0.17em}6.023}\times {\text{10}}^{\text{23}}"\text{O\hspace{0.17em}atoms}"$

Hence, $\text{Bulb\hspace{0.17em}X = Bulb Y}$