The correct option is B 8
A cross between AABBCC and aabbcc genotypes produces F1 hybrid with AaBbCc genotype. Total number of types of gamete produce by an organism is 2n, where n is the number of heterozygous genes present. The hybrid AaBbCc is heterozygous for three genes, thus total possible gametes by it = 23=8. The gametes produced by AaBbCc are "ABC, ABc, AbC, aBC, abC, aBc, abc and Abc". Correct answer is B.