Condition: $7A3$ is divisible by $3$.

The sum of its digits is $7+A+3=10+A$. For divisibility by $3$, $10+A$ must be divisible by $3$. Possible values of $A$ are $2$, $5$, or $8$. Equation: $136+5B7=7A3$.

Try Values of $A$: If $A=2$: $7A3=723$. $723-136=587$. $B=8$ (since $587 = 5B7$).

If $A = 5$ or $8$: The subtraction gives numbers that don’t match the form $5B7$.

The only possible value of $B$ is $8$.