Calculate relative lowering of vapor pressure of 1 molal glucose...

A $1$ molal solution of glucose means $1$ mole of glucose is dissolved in $1000$ g of water.

Number of moles of solute (glucose), $n = 1$ mol

Number of moles of solvent (water), $N = \dfrac{1000}{18} = 55.55$ mol

According to Raoult's law, the relative lowering of vapor pressure is equal to the mole fraction of the solute.

$\dfrac{\Delta P}{P^{{}^{\circ}}} = \chi_{\text{solute}} = \dfrac{n}{n + N}$

$\dfrac{\Delta P}{P^{{}^{\circ}}} = \dfrac{1}{1 + 55.55} = \dfrac{1}{56.55}$

$\dfrac{\Delta P}{P^{{}^{\circ}}} \approx 0.01768 \approx 1.8 \times 10^{-2}$

Answer: $1.8\times10^{-2}$