Answer
159.8 g of $Br_2$ has the greater number of atoms.
Work Step by Step
Molar mass $Br_2$ = $79.904 \times 2 = 159.808\ g$.
Therefore:
159.8 g of $Br_2$ has about 1 mol of $Br_2$ molecules.
$6.022 \times 10^{23}$ molecules of $Br_2$.
$6.022 \times 10^{23} Br_2 > 1\ Br_2$
Since we are comparing the same type of molecule ($Br_2$), the number of atoms will have the same result.
