First remove left recursion from the grammer
S -> AB
A -> Ca | $\epsilon$
B -> cB'
B' -> aACB' | $\epsilon$
C -> b | $\epsilon$
first ( S ) = { b, c, a }
first (A ) = {b, a, $\epsilon$ }
first (B) = {c}
first (C) = {b, $\epsilon$ }
follow (S) = { dollar }
follow (A) = {c, b , dollar, a}
follow (B) = { dollar , a} [a because first of B' is subset of follow of B]
follow (C) = {dollar , a}
Yes all are correct.