Principle of Carius method
In Carius method, halogen present in the organic compound is converted into silver halide.
Here, bromine is precipitated as $\mathrm{AgBr}$.

Molar mass of $\mathrm{AgBr}$ $M(\mathrm{AgBr}) = M(\mathrm{Ag}) + M(\mathrm{Br}) = 108 + 80 = 188$

Mass of bromine present in $0.15\,\text{g}$ of $\mathrm{AgBr}$ In $188\,\text{g}$ of $\mathrm{AgBr}$, mass of bromine $= 80\,\text{g}$.

Therefore, in $0.15\,\text{g}$ of $\mathrm{AgBr}$, $\text{mass of Br} = 0.15 \times \frac{80}{188}$ $= \frac{12}{188} = 0.06383\,\text{g}$

Percentage of bromine in the organic compound Mass of organic compound $= 0.25\,\text{g}$.

$\%\,\mathrm{Br} = \frac{0.06383}{0.25} \times 100$ $= 25.53\%$

Match with the required form The question asks for: $\underline{\hspace{2cm}} \times 10^{-1}\%$

Since $25.53\% = 255.3 \times 10^{-1}\%$

Nearest integer $= 255$.

Final answer $\boxed{255}$