Ratio of hybrid orbitals to pure ‘p’ orbitals in Ethene molecule is

${\mathrm{H}}_2\stackrel{{\mathrm{sp}}^2}{\mathrm{C}}=\stackrel{{\mathrm{sp}}^2}{\mathrm{C}}{\mathrm{H}}_2$

Each ‘sp²’ carbon contains 3 'sp²' hybrid orbitals and 1 pure ‘p’ orbital ;

H-atom utilizes its pure ‘s’ orbitals in bonding.

Number of 'sp²' hybrid orbitals = 2$\times$3 = 6

Number of pure ‘p’ orbitals     = 2$\times$1 = 2

Number of pure ‘s’ orbitals     = 4$\times$1 = 4

Ratio of hybrid orbitals to pure ‘p’ orbitals = 6 : 2 = 3 : 1 ;