Yes given solution is correct
Prefix for 'baba' starts from beginning so $\epsilon$ , 1 length = b, 2 length = ba , 3 length = bab,4 length = baba
postfix for 'baba' is counting from end but you have put each alphabet in stack and write it as it pop. For eg. string baba when put in stack for postfix, stack={a,b,a,b} and on pop, output is 'baba' (This is only for postfix evaluation).
so counting from end we got $\epsilon$, 1 length = a , 2 length = ba[stack={a,b}], 3 length = aba, 4 length = baba
thus intersection is {ba, baba, $\epsilon$}