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(B) The number of atoms is calculated as: $	ext{Number of atoms} = 	ext{Atomicity} 	imes 	ext{moles} 	imes N_A$.$(A)$ $Na_2CO_3$: $	ext{Atomicit

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$A, B$ and $D$ only

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(B) The number of atoms is calculated as: $	ext{Number of atoms} = 	ext{Atomicity} 	imes 	ext{moles} 	imes N_A$.$(A)$ $Na_2CO_3$: $	ext{Atomicity} = 6$,$	ext{moles} = \frac{212}{106} = 2$. $	ext{Atoms} = 6 	imes 2 	imes N_A = 12 N_A$.$(B)$ $Na_2O$: $	ext{Atomicity} = 3$,$	ext{moles} = \frac{248}{62} = 4$. $	ext{Atoms} = 3 	imes 4 	imes N_A = 12 N_A$.$(C)$ $NaOH$: $	ext{Atomicity} = 3$,$	ext{moles} = \frac{240}{40} = 6$. $	ext{Atoms} = 3 	imes 6 	imes N_A = 18 N_A$.$(D)$ $H_2$: $	ext{Atomicity} = 2$,$	ext{moles} = \frac{12}{2} = 6$. $	ext{Atoms} = 2 	imes 6 	imes N_A = 12 N_A$.$(E)$ $CO_2$: $	ext{Atomicity} = 3$,$	ext{moles} = \frac{220}{44} = 5$. $	ext{Atoms} = 3 	imes 5 	imes N_A = 15 N_A$.Thus,$A, B$ and $D$ have an equal number of atoms $(12 N_A)$.

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