Answer
$$\Delta H_{vap} = 30.7 \ kJ/mol$$
Work Step by Step
$ C_6H_6 $ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 6 )= 78.11 g/mol
$$q_{vap} = n \Delta H_{vap}$$
$$\Delta H_{vap} = \frac{q_{vap}}{n} = \frac{23.6 \ kJ}{60.0 \ g \times \frac{1 \ mol}{78.11 \ g}} = 30.7 \ kJ/mol$$
