Correct!

Amount of antigen used = 0.02 (ml = 20µl) x 4 (mg/ml BSA) = 0.08mg.

Concentration in cuvette = 0.04mg/ml (1ml precipitate + 1ml 0.1M NaOH).

If 1.9mg/ml gives an A₂₈₀ of 1.0, then A_280(Ag) = (0.04 x 1)/1.9 = 0.021.

Subtraction of this value from the total A₂₈₀ of the precipitate gives the A₂₈₀ for the antibody:

A_280(Ab) = A_280(ppt) - A_280(Ag)

Then, given that the A₂₈₀ for IgG is 1.0 at 0.7mg/ml, you can calculate the CONCENTRATION of antibody at equivalence (don't forget to take into account any dilutions made):

A)
A_280(Ab) = 0.265 - 0.021 = 0.244
0.244 x 0.7 = 0.17mg IgG.

B)
A_280(Ab) = 0.265 - 0.021 = 0.244
0.244 x 0.7 = 0.17mg/ml
0.17 x 2 = 0.34mg IgG.

C)
A_280(Ab) = 0.265 + 0.021 = 0.286
0.286 x 0.7 = 0.2mg/ml
0.2 x 2 = 0.4mg IgG.
D)
A_280(Ab) = 0.265 + 0.021 = 0.286
0.286 x 0.7 = 0.2mg/ml