According to Raoult's law, the partial vapour pressure of a component in a solution is given by $p_i = p_i^0 x_i$. For a $1:1$ molar mixture, the mole fractions of benzene and toluene in the liquid phase are equal ($x_{\text{benzene}} = x_{\text{toluene}} = 0.5$). Partial pressure of benzene, $p_{\text{benzene}} = 12.8 \text{ kPa} \times 0.5 = 6.4 \text{ kPa}$. Partial pressure of toluene, $p_{\text{toluene}} = 3.85 \text{ kPa} \times 0.5 = 1.925 \text{ kPa}$. Since the partial pressure of benzene is greater than that of toluene in the vapour phase ($p_{\text{benzene}} > p_{\text{toluene}}$), its mole fraction in the vapour phase ($y_i = \frac{p_i}{P_{\text{total}}}$) will also be higher. As a general rule, at equilibrium, the vapour phase is always richer in the component which is more volatile (i.e., has a higher vapour pressure in its pure state). Thus, the vapour will contain a higher percentage of benzene.