Router C has recived routing table from B, D and F.
Going via B gives (3 + 3, 0 + 3, 6 + 3, 8 + 3, 4 + 3, 2 + 3) = (6, 3, 9, 11, 7, 5)
Going via D gives ( 12 + 2, 13 + 2 + 2, 5 + 2, 0 + 2, 8 + 2, 10 + 2) = (14, 17, 7, 2, 10, 12)
Going via F gives (8 + 1, 9 + 1, 7 + 1, 4 + 1, 4 + 1, 0 + 1) = (9, 10, 8, 5, 5, 1)
Taking the minimum for each destination (A, B, C, D, E, F) except for C is (6, 3, 0, 2, 5, 1) via
(B, B, -, D, F, F).
Hence (b) is the correct option.