For given F1, F3 and F in canonical sum of product (SOP) form (in decimal) for the circuit, compute F2.…

In the given logic circuit, $F_1$ and $F_2$ feed an AND gate whose output is ORed with $F_3$, so in minterm-set notation $F = (F_1 \cap F_2) \cup F_3$. Because minterm $8$ is present in $F = \Sigma m(1, 6, 8, 15)$ but not in $F_3 = \Sigma m(1, 6, 15)$, it must belong to $F_1 \cap F_2$, so $8 \in F_2$. Conversely, minterm $4 \in F_1$ is absent from $F$, so $4 \notin F_2$ (ruling out $\Sigma m(4, 6)$, $\Sigma m(4, 8)$, and $\Sigma m(4, 6, 8)$). Substituting $F_2 = \Sigma m(6, 8)$ gives $(F_1 \cap F_2) \cup F_3 = \{6, 8\} \cup \{1, 6, 15\} = \Sigma m(1, 6, 8, 15) = F$.