The teacher goes round the circle four times. If there was an even number of children, the sixth child would always receive a blue present and would have four blue presents at the end. Since they only have two blue presents, there must be an odd number of children in the circle.
With the same logic for the green presents the answer can’t be a multiple of 3.
The sixth child receives a green present in the first round because 6 is a multiple of 3. She only receives one more green present so the answer can’t be a multiple of 3.
The options are 7, 11, 13, 17, 19 … and so 7 is the fewest number of children.