Hi. Actually, the problem requires no special equations. All you need is the ideal gas law.
The only relevant information is the volume of air, the R constant, the temperature, and the ambient pressure to calculate the total moles of air molecules bubbled through the benzene. With that, and the 1.705 g of benzene (C6H6: MM = 78 g/mol), you can calculate the vapor pressure of benzene afterwards:
n_air = P * V_air / RT = .198 moles
n_benzene = m_benzene / MM_benzene = 1.705 / 78 = .021 moles
P_benzene = P_air * n_benzene / n_air = 100.56 * .021 / .198 = 10 kPa