The proportion of the offspring with AABBCC genotype in a cross between two AaBbCc individuals will be 1/64.
Individuals with AaBbCc genotype will produce the following gametes: ABC, ABc, AbC, Abc, aBC, aBc, abC, and abc.
The male and female gametes will randomly unite during fertilization to produce 64 offspring according to the attached Punnet's square.
Thus, only 1 out of the 64 possible offspring can have the genotype AABBCC. In other words, the proportion of the offspring to have AABBCC genotype would be 1/64.
More on genetic probabilities can be found here: https://brainly.com/question/851793?referrer=searchResults
