$= 4 \times 6.023 \times 10^{23}$ atoms 

$71.0\,gm$ of chlorine $ = \frac{{71.0}}{{71.0}} = 1$ moles of chlorine

                                    $= 6.023 \times 10^{23}$ atoms

$127\,gm$ of iodine $ = \frac{1}{2}$ mole of $I_2$

$ = \frac{{127}}{{254}} = 6.023 \times {10^{23}} \times \frac{1}{2}$ atoms

$48.0\,gm$ of magnesium

$ = \frac{{48.0}}{{24.0}} = 2 \times 6.023 \times {10^{23}}$ atoms

$\therefore $ $4.0\,gm\,H_2$ has largest number of atoms