1. For a first-order reaction, $t = \frac{2.303}{k}\log\frac{[A]_0}{[A]}$

2. For $67\%$ completion, the reactant left is $33\%$: $\frac{[A]}{[A]_0} = 0.33 = \frac{33}{100}$ Hence, $t_{67\%} = \frac{2.303}{k}\log\frac{100}{33}$

3. The half-life for a first-order reaction is: $t_{1/2} = \frac{0.693}{k} = \frac{2.303}{k}\log 2$

4. Therefore, $\frac{t_{67\%}}{t_{1/2}} = \frac{\log(100/33)}{\log 2}$

5. Now evaluate $\log(100/33)$: $\log\frac{100}{33} = \log 100 - \log 33$ $= 2 - (\log 3 + \log 11)$ This is awkward directly, so write: $\frac{100}{33} = 3.03\ldots$ More neatly, $\log\frac{100}{33} = \log 100 - \log 33 = 2 - (\log 3 + \log 11)$

   But the standard shortcut is: $\frac{100}{33} \approx 3.03$ so $\log\frac{100}{33} \approx \log 3.03 \approx 0.481$

   Also, $\log 2 = 0.3010$

   Hence, $\frac{t_{67\%}}{t_{1/2}} \approx \frac{0.481}{0.301} \approx 1.60$

6. Given that $t_{67\%} = x \times 10^{-1} \times t_{1/2}$ so $x \times 10^{-1} = 1.60$ $x = 16.0$

7. Nearest integer: $\boxed{16}$

The derived answer matches the stored correct answer.