Principle of sulphur estimation
In Carius method, sulphur present in the organic compound is converted into sulphate and finally precipitated as barium sulphate, $\text{BaSO}_4$.

Molar mass of $\text{BaSO}_4$
Given atomic masses:
$\text{Ba} = 137$, $\text{S} = 32$, $\text{O} = 16$

$M(\text{BaSO}_4)=137+32+4\times 16=137+32+64=233$

Mass of sulphur in $1.44\text{ g}$ of $\text{BaSO}_4$
In $233\text{ g}$ of $\text{BaSO}_4$, mass of sulphur $=32\text{ g}$.

Therefore, in $1.44\text{ g}$ of $\text{BaSO}_4$, $\text{Mass of S} = \frac{32}{233}\times 1.44$

$\text{Mass of S} = \frac{46.08}{233} \approx 0.1978\text{ g}$

Percentage of sulphur in the compound
Mass of organic compound $=0.471\text{ g}$

$\%\text{S} = \frac{0.1978}{0.471}\times 100$

$\%\text{S} \approx 42.0\%$

Nearest integer
$\boxed{42}$