Answer
$136torr$
Work Step by Step
We know that
moles of $C_3H_8O_3=53.6g\times \frac{1mol}{92.09g}=0.582mol\space C_3H_8O_3$
moles of $C_2H_5OH=133.7g\times \frac{1mol}{46.07g}=2.90mol\space C_2H_5OH$
Now total moles $=0.582+2.90=3.48mol$
Using Raoult's law, we have
$113 torr=\frac{2.90mol}{3.48mol}\times P^{\circ}_{C_2H_5OH}$
This simplifies to:
$P^{\circ}_{C_2H_5OH}=136torr$
