There are $3$ ways to assign B to a room, and then $2$ ways to assign D. That leaves the pair AC and the individuals E and F to be assigned; each of these three ‘people’ can be assigned to any of the three rooms, so they can be assigned in $3^3$ ways. The total number of allowable assignments is therefore $$3\cdot2\cdot3^3=2\cdot3^4=162\;.$$
Note that your answer can’t possibly be right: if there were no restrictions, each of the $6$ people could be assigned to any of the $3$ rooms; that’s a $3$-way choice made $6$ times, so it can be done in $3^6=729$ ways. Thus, your answer is way bigger than the number of possible assignments when there are no restrictions at all. The number of assignments satisfying the restrictions on A, B, C, and D must be smaller than $729$.