This is a permutation and combination problem.
If the element existed as Z, and not $Z_3$, then yes you are right the peaks should be 2:1, but more correctly abundance would be 66%:33% rather than 100%:50%. Keep this in mind, it will be helpful when you do the calculation.
But now we are told that it exists as $Z_3$, which means we have to have 3 atoms of Z.
There is only one "way" to get 171, and that is to have all three atoms as 57Z. So from probability, this would look like $$P(M=171)=(1/3)^3 = 1/27$$
Similarly, we consider the case of getting 165, which is all three atoms have to be 55Z. Mathematically, this would then be $$P(M=165)=(2/3)^3 = 8/27$$
In the case of M=167, we need two 55Z's and one 57Z. So after doing the math we find that it is $4/27$.
Since this is homework, I'm sure you can do the case of M=169 by yourself now.