In an estimation of sulphur by Carius method 0.2 g of the substance...

Mass of the organic compound, $W = 0.2$ g

Mass of $\text{BaSO}_4$ formed, $m = 0.6$ g

Molar mass of $\text{BaSO}_4 = 233 \text{ g mol}^{-1}$

Molar mass of $\text{S} = 32 \text{ g mol}^{-1}$

The percentage of sulphur in the organic compound is given by the formula:

$\% \text{S} = \dfrac{32}{233} \times \dfrac{m}{W} \times 100$

Substituting the given values:

$\% \text{S} = \dfrac{32}{233} \times \dfrac{0.6}{0.2} \times 100$

$\% \text{S} = \dfrac{32}{233} \times 3 \times 100$

$\% \text{S} = \dfrac{9600}{233} \approx 41.20\%$

Rounding off to the nearest integer, the percentage of sulphur is $41$.

Answer: $41$