smc

We can expand the fraction $0.{\overline{23}}_k$ as follows: $0.{\overline{23}}_k=2\cdot k^{-1}+3\cdot k^{-2}+2\cdot k^{-3}+3\cdot k^{-4}+\cdots$ Notice that this is equivalent to $$2(k^{-1}+k^{-3}+k^{-5}+...)+3(k^{-2}+k^{-4}+k^{-6}+\cdots )$$ By summing the geometric series and simplifying, we have $\frac{2k+3}{k^2-1}=\frac{7}{51}$. Solving this quadratic equation (or simply testing the answer choices) yields the answer $k=\boxed{16}$.