18.4 g of glycerol, with molar mass 92.0 g/mol, is dissolved in 90.0 g of water at 298 K…

Step 1. Moles. $n_B = \dfrac{18.4\ \text{g}}{92.0\ \text{g/mol}} = 0.200\ \text{mol}$ and $n_A = \dfrac{90.0\ \text{g}}{18.0\ \text{g/mol}} = 5.00\ \text{mol}$.

Step 2. Mole fraction of the solute. $x_B = \dfrac{0.200\ \text{mol}}{5.00\ \text{mol} + 0.200\ \text{mol}} = 0.0385$.

Step 3. The lowering. $p^\circ - p = x_B\,p^\circ = 0.0385 \times 3.17\ \text{kPa} = 0.122\ \text{kPa}$.

Step 4. The vapour pressure. $p = 3.17\ \text{kPa} - 0.122\ \text{kPa} = 3.048\ \text{kPa}$, which is $3.05\ \text{kPa}$.