If you are ranking $13$ cards of each suit from $1$ to $13$, there are $7$ odd number cards in each suit. So in total there are $28$ odd number cards and $24$ even number cards.

Hence number of hands of $6$ cards where there are at least $3$ odd number cards is be given by,

$\displaystyle {28 \choose 3} {24 \choose 3} + {28 \choose 4} {24 \choose 2} + {28 \choose 5} {24 \choose 1} + {28 \choose 6}$

First on your working:

If you are ranking $13$ cards of each suit from $1$ to $13$, there are $7$ odd number cards in each suit. So in total there are $28$ odd number cards and $24$ even number cards.

Hence number of hands of $6$ cards where there are at least $3$ odd number cards is be given by,

$\displaystyle {28 \choose 3} {24 \choose 3} + {28 \choose 4} {24 \choose 2} + {28 \choose 5} {24 \choose 1} + {28 \choose 6}$

But if the question considers odd number cards as $1, 3, 5, 7, 9$ as in a usual deck (Ace considered $1$), then there are $20$ odd number cards and $32$ of rest. You can apply the same logic as earlier to calculat number of hands with at least $3$ odd number cards.