jmc

Because $b<10^b$ for all $b>0$, it follows that ${\log }_{10}b<b$. If $b\ge 1$, then $0<\left({\log }_{10}b\right)/b^2<1$, so $a$ cannot be an integer. Therefore $0<b<1$, so ${\log }_{10}b<0$ and $a=\left({\log }_{10}b\right)/b^2<0$. Thus $a<0<b<1<1/b$, and the median of the set is $\boxed{b}$.

Note that the conditions of the problem can be met with $b=0.1$ and $a=-100$.