Given,

Vapour pressure of benzene, ${\mathrm{P}}_{\mathrm{B}}^{°}=50 \mathrm{mmHg}$

Vapour pressure of toluene, ${\mathrm{P}}_{\mathrm{T}}^{°}=40 \mathrm{mmHg}$

Molar mass of benzene $=78 \mathrm{g} {\mathrm{mol}}^{-1}$

Molar mass of toluene $=92 \mathrm{g} {\mathrm{mol}}^{-1}$

Moles of benzene $=\frac{117}{78}=1.5$

Moles of toluene $=\frac{46}{90}=0.51$

Mole fraction of benzene $=\frac{1.5}{1.5+0.511}=0.75$

Mole fraction of toluene $=\frac{0.511}{1.5+0.511}=0.25$

${\mathrm{P}}_{\mathrm{B}}={\mathrm{P}}_{\mathrm{B}}^{°}{\mathrm{X}}_{\mathrm{B}}=50\times 1.5=75 \mathrm{mmHg}$

${\mathrm{P}}_{\mathrm{T}}={\mathrm{P}}_{\mathrm{T}}^{°}{\mathrm{X}}_{\mathrm{T}}=40\times 0.511=20.44 \mathrm{mmHg}$

Total vapour pressure $=75+20.44=95.44 \mathrm{mmHg}$

Mole fraction of benzene in vapour phase $=\frac{20.44}{95.44}=0.21$