Answer: D

Solution:

The repeating base-$k$ fraction is $2k^{-1}+3k^{-2}$ $+2k^{-3}+3k^{-4}+\cdots$. Grouping odd and even powers gives $$\begin{aligned} &2(k^{-1}+k^{-3}+\cdots) \\ &\quad {}+3(k^{-2}+k^{-4}+\cdots). \end{aligned}$$

Using geometric series, these sums are $\dfrac{2k}{k^2-1}$ and $\dfrac{3}{k^2-1}$, so $0.\overline{23}_k=\dfrac{2k+3}{k^2-1}$.

Setting $\dfrac{2k+3}{k^2-1}=\dfrac{7}{51}$ gives $51(2k+3)=7(k^2-1)$, so $7k^2-102k-160=0$. Hence $k=16$. Thus, D is the correct answer.