I've done in this way.
For selecting 2 pairs, first, we choose 2 types of cards out of possible 13. So, we get 13C2. Now, for each pair, we have to select 2 colours out of 4. So, now, we get 13C2 * 4C2 * 4C2, as final for selecting the pairs.
But, now for selecting the last card, we will select 1 card from rest 11 numbers, which will be 11C1, but it can be of any 4 colours, so for single card, the number of combinations are 11C1 * 4C1.
Finally, answer :
13C2 * 4C2 * 4C2 * 11C1 * 4C1 / 52C5 = 0.0475