The probability that a dihybrid cross of heterozygotes will produce an individual exhibiting BOTH recessive traits is:
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The best way to do a question like this is to set up a dihybrid cross and do it. It tells you that both parents are heterozygotes for both traits. So you could call their genotpyes AaBb. What you want to do next is set up the actual Punnett Square for AaBb X AaBb, and count the number of offspring that are AaBb themselves.
Is that enough guidance, or would you like help setting up the Punnett Square?
I think it is b, right?
Oh, I read the question wrong. It wants the offspring that will be aabb.
What you should do is actually set up the square, fill in the offspring's genotypes, and see how many of them are aabb.