The ratio of hybrid and unhybrid orbitals involved in the bonding of a benzene molecule is

Number of hybrid orbitals = number of carbons $\times$ number denoting the hybridization of carbon

= 6  $\times$ 3 =18

Number of pure orbitals = number of hydrogen present $+2\times$(number of $\pi$ - bonds present)

 = 6 +2 $\times$(3) =12

Ratio of hybrid and unhybrid orbitals =18:12= 3:2