Here is my thought process:
Tube 6 had the addition of CuSO4, according to Le Chat's principle, this will allow rxn 1 to shift toward the left side (because there is an increase of SO4-2 ions from CuSO4), causing an increase in I- ions. Relating this to rxn 2, an increased amount of I- will cause a shift to the left causing S4O6 to be used up.
I definitely overthought about this because it was simply, overtime, S4O6 gets used up, so B is the correct answer.
Maybe I didn't understand the passage? Why would D be incorrect?