$\text { Given } \lambda_{1}=\frac{\ell {n} 2}{700} / \text { year }, \lambda_{2}=\frac{\ell {n} 2}{1400} / \text { year }$

$\therefore \lambda_{\text {net }}=\lambda_{1}+\lambda_{2}=\ell n 2\left[\frac{1}{700}+\frac{1}{1400}\right]=\frac{3 \ell {n} 2}{1400} / \text { year }$

Now, Let initial no. of radioactive nuclei be No

$\therefore \frac{{N}_{0}}{3}={N}_{0} {e}^{-\lambda_{{mt}} {t}}$

$\Rightarrow \ell {n} \frac{1}{3}=-\lambda_{\text {net }} {t}$

$\Rightarrow 1.1 \frac{3 \times 0.693}{1400} {t} \Rightarrow {t} \approx 740 \text { year }$