If the total income of a family per month is $6000

As it is given that the middle value or median is $2000, hence two cases are possible:
i) When there are even number of members.
In such case, answer choices A, C and E are being considered.
Since for sets consisting of even number of elements, the mean of middle two values is the mean. Hence ATLEAST these two values must sum to $4000. The largest value has to be atleast $2000 and the lowest value has to be atmost $2000. Thus sum would be more than $6000. Hence C and E are eliminated. For answer choice A, where there are 2 elements, the mean of these two elements must be $2000 or sum is $4000. Eliminate A too.
Answer choice B is the only possible answer, in which either all the three elements are the same or the largest is more than $2000 and the smallest is less than $2000.
In case of E, the sum would become more than $6000.
+1B