The ratio between hybridized and pure orbitals in o-xylene are.

Let us draw the structure of o-xylene as follows and observe the number of hybridized carbons first.

1,2,3,4,5,6 Carbons, Are $s{p}^2$ hybridization,  Due to which we can say that around each carbon they are three sigma bonds,$\text{'}C\text{'}3\sigma$  bonds. Carbon number 7 and 8 carbons are $s{p}^3$ hybridization, Therefore around them all are sigma bonds ‘C’ $4\sigma$.
Calculation:
Each $s{p}^2$  carbon have 3 hybrid orbitals
Each  $s{p}^3$  carbon have 4 hybrid orbitals
Each  $s{p}^3$ carbon have zero pure orbitals
All hydrogen’s have pure orbitals
Total hybrid orbitals:  $s{p}^2=3\times 6=18$

                                   $s{p}^3=4\times 2=8$

                                 ------------------------                                              
                                                   = 26
Total pure orbitals from the 6 $s{p}^2$ carbons are (one p-orbital is pure from each carbon) = 6
                       All hydrogens are pure, 10 Hydrogens are present =10 
                                                    Total pure orbitals are=16
Now ratio:  $\frac{\cancel{26}}{\cancel{16}}=\frac{13}{8}=13:8$