jmc

First, we have $$53_6=5\cdot 6^1+3\cdot 6^0=33_{10}.$$ and $$113_b=1\cdot b^2+1\cdot b^1+3\cdot b^0=(b^2+b+3{)}_{10}.$$ Therefore, we must have $b^2+b+3=33$, so $b^2+b-30=0$. Factoring we have $(b-5)(b+6)=0$. Thus, $b=5$ or $b=-6$. The positive value is $b=\boxed{5}$.