First observe that the two adults may not sit in the same row. Since there are three seats in each row, there are 3 × 3 = 9 ways of selecting two seats for the adults to sit. If the two adults sit in the two center seats in each row, then the four children can sit in any of the other four seats. There are 2 ways for the adults to seat themselves in the two central seats, and there are 4! = 24 ways for the children to sit in the four other seats. Thus, there are 2 × 24 = 48 ways for the people to sit with the adults in the two center seats.

If either adult sits in an end seat, then there are two seats next to that adult where either two little boys ortwo little girls will need to sit. Thus, the little boys will need to sit in one row, and the little girls will need to sit in the other row. There are 9 − 1 = 8 ways to choose the pair of seats for the adults to sit so that atleast one adult sits in an end seat. There are two ways for the adults to choose which of the two seats to sit in, so there are 2 × 8 = 16 ways for the adults to choose seats. There are two ways for the little boys to choose a row to sit in. Then there are 2 × 2 ways for the little boys and little girls to choose seats. This accounts for 16 × 2 × 2 × 2 = 128 ways for the six people to sit. Thus, the final count is 48 + 128 = 176.

Hence, option 4.