The geneticists studying the alien organisms have deciphered their genetic code. For Familiarity, they have used A, C, G, and T to indicate the bases in this code. A comparable genetic sequence in each alien has been identified in the genes that code for eye growth. This genetic sequence has allowed the aliens to be divided into six groups, with the single-tail three-eye group split in half. The geneticists also studied several outgroup species (not shown) to compile a likely ancestral DNA sequence. From this information, they have constructed four possible phylogenetic trees. Based on this genetic information, which phylogenetic tree is most parsimonious?
a) phylogenetic tree I
b) phylogenetic tree II
c) phylogenetic tree III
d) phylogenetic tree IV

Your question is incomplete, but I attached the exercise with the possible phylogenetic trees and all the information that your question missed. I hope it helps you.

The most parsimoniuous phylogenetic tree is c) phylogenetic tree III

Explanation:

The most parsimonious tree is the one that takes fewer base-change events, it's the simpler one. In this case it could be tree III or tree IV. But, if we look at the genetic evidence, group 2b and 3 are nearer than 2b and 2a; group 4 and 5 are also closely related. So it is possible that the group 2b, as well as groups 3,4 and 5, diverged from group 2a. So the answer is tree III.