smc

Converting everything into base $10$, we have $(b+2)(b+5)(b+6)=3b^3+b^2+4b+6$. Looking ahead, the constant term of the polynomial will be $2\ast 5\ast 6-6=54$. By the Rational Root Theorem, the only possible integer roots are $1,2,3,6,9,18,27,54$. Bases $1,2,3,6$ do not have a $6$ as a digit. Testing $b=9$ gives a solution that works. Therefore, we are working in base $9$. Adding the units place in base $9$, $2+5+6=14_9$, so we carry the $1$ to get a total of $44_9$, which is option $\boxed{44}$.