= Solution
\b[$\beta=-2.4\ \mathrm{eV}$ and $\alpha=-6.8\ \mathrm{eV}$.] The <Hückel method> gives benzene's occupied frontier level as $\epsilon_H=\alpha+\beta$ and its unoccupied frontier level as $\epsilon_L=\alpha-\beta$. Therefore
$$
4.8=\epsilon_L-\epsilon_H=-2\beta.
$$
Using the <Koopmans' theorem> approximation for the <ionization energy>, $I\simeq-\epsilon_H=9.2$, gives $\alpha+\beta=-9.2$ and the stated parameters. This calibration neglects orbital relaxation, electron correlation and any distinction between a measured optical excitation and an independent-particle orbital gap.
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