$$ \lambda t=\ln \frac{\left({ }^{14} \mathrm{C} /{ }^{12} \mathrm{C}\right)_{\text {atmosphere }}}{\left({ }^{14} \mathrm{C} /{ }^{12} \mathrm{C}\right)_{\text {wood sample }}} $$ As per the question, $$ \frac{\left({ }^{14} \mathrm{C} /{ }^{12} \mathrm{C}\right)_{\text {wood }}}{\left({ }^{14} \mathrm{C} /{ }^{12} \mathrm{C}\right)_{\text {atmosphere }}}=\frac{1}{8} $$ So, $\lambda \mathrm{t}=\ln 8$ $$ \begin{aligned} & \frac{\ln 2}{\mathrm{t}_{1 / 2}} \mathrm{t}=\ln 8 \\ & t=3 \times t_{1 / 2}=17190 \text { years } \end{aligned} $$