$ = \frac{{0.16\,g}}{{32\,g/mol}} = 0.005\,mole$

$2NaCl{O_3}\xrightarrow{\Delta }2NaCl + 3{O_2}$

According to the reaction, 

$3$ moles of $O_2$ $=$ $2$ moles of $NaCl$ $=$ $2$ moles of $AgCl$

Molar mass of $AgCl = 143.5\,g/mol$

$0.005$ moles of $O_2$ will ppt. $ = 0.005 \times \frac{2}{3}$ moles $AgCl$

$= 0.0033$ moles of $AgCl$

$\therefore $ Mass of $AgCl$ (in $g$) obtained will be

$= 143.5\,g/mol \times 0.0033$ $moles$ $= 0.48\,g$