How to determine the mole fraction of the liquid phase from a given mole fraction of the vapour phase for a benzene/toluene mixture?
The question gives you the mol percentage of benzene ($53 \%$) in the vapour phase, and not in the liquid phase.

It's always good to write down what you have first. (note: I use $y$ to denote mole fractions in the vapour phase and $x$ to denote mole fractions in the liquid state)

First of all, we have $y_\text{benzene} = 0.53$. Additionally, we have been given the vapour pressure of pure benzene and toluene (denoted with a superscript $*$):

$$p^*_\text{benzene} = 0.125\ \mathrm{atm}  \qquad p^*_\text{toluene} = 0.037\ \mathrm{atm} $$

and we require $x_\text{benzene}$

We also, know $p_\text{benzene} = x_\text{benzene}\ p^*_\text{benzene}$, and the total vapor pressure is given by $$P = p_\text{benzene} + p_\text{toluene} = x_\text{benzene}\ p^*_\text{benzene} + x_\text{toluene}\ p^*_\text{toluene} = x_\text{benzene}\ p^*_\text{benzene}+ (1- x_\text{benzene})\ p^*_\text{toluene}$$

Finally, 

$$y_\text{benzene} = \frac{p_\text{benzene}}{P} = \frac{x_\text{benzene}\ p^*_\text{benzene}}{x_\text{benzene}\ p^*_\text{benzene}+ (1- x_\text{benzene})\ p^*_\text{toluene}}$$

It is now, trivial to solve this equation. Since I am lazy, and was not in the mood to write down two lines of algebra and reach for a calculator, I chucked it into Mathematica, and the result is: 



Thus, $x_\text{benzene} = 0.25$, which is indeed the result given in your answer key.