To solve this problem, we can break it down into different cases based on the arrangement of the pairs of digits.

Case 1: Two pairs of the same digit
In this case, we have two pairs of the same digit and two other digits. We can choose the two digits to form the pairs in ${4 \choose 2} = 6$ ways. The remaining two digits can be arranged in $2! = 2$ ways. Therefore, there are a total of $6 \times 2 = 12$ 6-digit numbers in this case.

Case 2: Two pairs of different digits
In this case, we have two pairs of different digits and two other digits. We can choose the two pairs of digits in ${4 \choose 2} \times {2 \choose 2} = 6$ ways. The remaining two digits can be arranged in $2! = 2$ ways. Therefore, there are a total of $6 \times 2 = 12$ 6-digit numbers in this case.

Case 3: One pair and two individual digits
In this case, we have one pair of digits and two individual digits. We can choose the pair of digits in ${4 \choose 2} = 6$ ways. The remaining two digits can be arranged in $2! = 2$ ways. Therefore, there are a total of $6 \times 2 = 12$ 6-digit numbers in this case.

Case 4: Two pairs of the same digit and one individual digit
In this case, we have two pairs of the same digit and one individual digit. We can choose the digit for the individual position in 4 ways. The remaining two digits can be arranged in $2! = 2$ ways. Therefore, there are a total of $4 \times 2 = 8$ 6-digit numbers in this case.

Total number of 6-digit numbers with exactly two pairs of digits
Adding up the numbers from all the cases, we get a total of $12 + 12 + 12 + 8 = 44$ 6-digit numbers.

Therefore, the correct answer is option C) 1080.