Which one of the following will have largest number of atoms?

The correct option is B $1\text{g Li}$
(i) $1\text{g Au}=\frac{1}{197}\text{mol atom of Au}$

$=\frac{1}{197}\times 6.022\times {10}^{23}\text{atoms of Au}$

(ii) $1\text{g Na}=\frac{1}{23}\text{mol atom of Na}$

$=\frac{1}{23}\times 6.022\times {10}^{23}\text{atoms of Na}$

(iii) $1\text{g Li}=\frac{1}{7}\text{mol atom of Li}$

$=\frac{1}{7}\times 6.022\times {10}^{23}\text{atoms of Li}$

(iv) $1\text{g}{\text{Cl}}_2=\frac{1}{71}\text{mol molecules of}{\text{Cl}}_2$

$=\frac{1}{71}\times 6.022\times {10}^{23}\text{molecules of}{\text{Cl}}_2$

$=\frac{2}{71}\times 6.022\times {10}^{23}\text{atoms of}{\text{Cl}}_2$

Hence, $1\text{g}$ lithium has the largest number of atoms.