Mass of $A g B r=141 \mathrm{mg}=0.141 \mathrm{g}$

$1\; mole$ of $A g B r=1\; g$ atom of $B r$

$188 g$ of $A g B r=80 g$ of $B r$

$188 g$ of $A g B r$ contain bromine $=80 g$

$0.141 g$ of $A g B r$ contain bromine $=\frac{80}{188} \times 0.141$

This much amount of bromine present in $0.250 \mathrm{g}$ of organic compound

$\because \%$ of bromine $=\frac{80}{188} \times \frac{0.414}{0.250} \times 100=24 \%$