The F 1 is all BbCc, so all individuals are brown. There are no gray or albino individuals. In the backcross to the bbcc population, half of the offspring are cc and hence albino, regardless of the genotype at the other locus. To lack C, an individual must be homozygous for cc.
Individuals are black if the dominant allele is present at both loci and brown otherwise.
You cross two pure lines to form an F 1 and then perform a testcross with a population homozygous for a and b. You want to test Mendel's second law. You do not need to calculate chi square. You do not need to carry out the tests, just describe them. You would do a chi square test with 3 degrees of freedom.