[Atomic mass : Silver $=108$, Bromine $=80]$
$\Rightarrow \quad$ Mass of $\mathrm{Br}=\frac{0.2397}{188} \times 80\, \mathrm{~g}$
therefore $\%$ Br in the organic compound
$=\frac{W_{\text {Br }}}{W_{T}} \times 100$
$=\frac{0.2397 \times 80}{188 \times 0.15} \times 100=0.85 \times 80$
$=68$
$\Rightarrow$ Nearest integer is $'68^{\prime}$
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