If $ a$ is a nonzero integer and $ b$ is a positive number such that $ ab^{2} \equal{} \log_{10}b,$ what is the median of the set $ \{0,1,a,b,1/b\}$?

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ a \qquad \textbf{(D)}\ b \qquad \textbf{(E)}\ \frac {1}{b}$
1. Given the equation \( ab^2 = \log_{10} b \), we need to find the median of the set \(\{0, 1, a, b, \frac{1}{b}\}\).
2. Let's start by solving the equation \( ab^2 = \log_{10} b \). We can rewrite \(\log_{10} b\) as \( \log_{10} b = x \), which implies \( b = 10^x \).
3. Substituting \( b = 10^x \) into the equation \( ab^2 = \log_{10} b \), we get:
   \[
   a (10^x)^2 = x \implies a \cdot 10^{2x} = x
   \]
4. To find a solution, we can try specific values for \(a\) and \(b\). Let's pick \(a = -100\) and \(b = 0.1\):
   \[
   b = 0.1 \implies \log_{10} 0.1 = -1
   \]
   \[
   ab^2 = -100 \cdot (0.1)^2 = -100 \cdot 0.01 = -1
   \]
   This satisfies the equation \( ab^2 = \log_{10} b \).
5. Now, we have \(a = -100\), \(b = 0.1\), and \(\frac{1}{b} = 10\). The set is \(\{0, 1, -100, 0.1, 10\}\).
6. Arranging the set in ascending order, we get:
   \[
   \{-100, 0, 0.1, 1, 10\}
   \]
7. The median of this set is the middle value, which is \(0.1\).

The final answer is \(\boxed{b}\).