A tri-hybrid cross is made between two yeasts, both with genotypes Aa Bb Cc. What proportion of the offspring will be of genotype aabbcc?

The correct option is C 1/64
AaBbCc x AaBbCc
The gamates from both the parents will be 8 in number as determined by formula 2^n (Here n = 3), So, its 8.

Gamates will be, ABC, ABc, AbC, Abc, abc, aBC, abC, aBc.
There is only one possibility that offspring will be of genotype aabbcc, i.e. when abc gamates from both the parents will combine. Total offspring will be 8 x 8 = 64.
So 1 out of 64 (1/64) will be of genotype aabbcc.