We have to get sum $13$ in $4$ ways.
- $A + B + C = 13$
- $C + D + E = 13$
- $E + F + G = 13$
- $G + H + I = 13$
Adding all $4$ we get
$A + B + D + F + H + I + 2(C + E+ G) = 52$
Since $A \to G$ sum to $45$ we get
$C + E + G = 52-45 =7$
Only option we can add to $7$ with $3$ distinct positive integers is $1 +2+4.$ So, $C,E,G \in \{1,2,4\}$
We also have $C+D+E = 13.$ If $C$ and $E$ are $1,2$ this is not possible as this requires $D$ to be $10$ which is not allowed. So, at least one of $E$ or $C$ must be $4$ also making $G$ to be either $1$ or $2.$
Since, $E + F + G = 13, E + G $ must also be $\geq4$ meaning $E$ and $G$ cannot be $1,2.$ Since, $G \leq 2,$ it means $E > 2.$
So, only option is $E = 4.$
Correct Option: B.