A substance ' X ' (1.5 g) dissolved in 150 g of a solvent ' Y '...

Using boiling point elevation to find molality:
$\Delta T_b = K_b \times m$ gives $0.5 = 5.0 \times m$, so $m = 0.1$ mol/kg.
Converting molality to moles of solute:
$0.1 = \frac{\text{moles of X}}{0.15}$, giving moles of X = 0.015 mol.
Molar mass of X = 1.5 g / 0.015 mol = 100 g/mol.
For vapor pressure lowering, the relative lowering equals the mole fraction of solute.
Moles of solvent Y = 150 g / 300 g/mol = 0.5 mol.
Relative lowering of vapor pressure
= $\frac{n_X}{n_X + n_Y} = \frac{0.015}{0.015 + 0.5}$
$= \frac{0.015}{0.515} = 0.0291 \approx 3 \times 10^{-2}$