Answer
$\displaystyle P[F_j|E]=\frac{P(F_j|E)P(E)}{\sum ^n_{i=1} P(E|F_i)P(F_i)}$
Work Step by Step
Observe that $P(E)=P(E\cap S)=P(E \cap (\cup ^n_{i=1}F_i))=P(\cup ^n_{i=1} E\cap F_i)=\sum ^n_{i=1} P(E\cap F_i),$ (the last step is because $F_1,F_2,..., F_n$ are mutually exclusive.
Thus,
$\displaystyle P[F_j|E]=\frac{P(F_j\cap E)}{P(E)}=\frac{P(F_j|E)P(E)}{\sum ^n_{i=1} P(E\cap F_i)}=\frac{P(F_j|E)P(E)}{\sum ^n_{i=1} P(E|F_i)P(F_i)}$
