For two radioactive elements $R_1$ and $R_2$, decay constants,
$\lambda_1=6 \lambda$ and $\lambda_2=\lambda$
Half-life of $R_2,\left(t_{1 / 2}\right)_2=1.4 \times 10^{17} \mathrm{~s}$
If $N_0$ be the initial number of nuclei at $t=0$, in both $R_1$ and $R_2$ or $N_1$ and $N_2$ be the nuclei after time $t$.
Then,
$N_1=N_0\left(\frac{1}{2}\right)^{n_1}$
and
$N_2=N_0\left(\frac{1}{2}\right)^{n_2}$
$\therefore \quad \frac{N_2}{N_1}=\left(\frac{1}{2}\right)^{n_2-n_1}$
$e=\left(\frac{1}{2}\right)^{n_2-n_1} \quad\left[\because \frac{N_2}{N_1}=e\right]$
$\begin{aligned} \log _e e & =\left(n_2-n_1\right) \log \frac{1}{2} \\ \Rightarrow \quad 1 & =\left(n_2-n_1\right) \log 2^{-1} \\ 1 & =\left(n_1-n_2\right) \log 2 \\ 1 & =\left(n_1-n_2\right) 0.7 \\ \Rightarrow \quad n_1-n_2 & =\frac{1}{0.7} \\ \frac{t}{\left(t_{1 / 2}\right)_1} & -\frac{t}{\left(t_{1 / 2}\right)_2}=\frac{1}{0.7}\end{aligned}$
$\begin{aligned} & \text { Since, }\left(t_{1 / 2}\right)_2=\frac{0.693}{\lambda_2}=\frac{0.693}{\lambda} \\ & \Rightarrow \quad \lambda=\frac{0.693}{\left(t_{1 / 2}\right)_2}=\frac{0.693}{1.4 \times 10^{17}}=0.495 \times 10^{-17} \\ & \therefore \quad \lambda_1=6 \lambda=6 \times 0.495 \times 10^{-17} \\ & =2.97 \times 10^{-17} \mathrm{~m} \\ & \therefore \quad\left(t_{1 / 2}\right)_1=\frac{0.693}{\lambda_1}=\frac{0.693}{2.97 \times 10^{-17}} \\ & \left(t_{1 / 2}\right)_1=0.23 \times 10^{17} \mathrm{~s} \\ & \end{aligned}$
$\therefore$ From Eq. (i), we get
$\begin{aligned} \frac{t}{0.23 \times 10^{17}}-\frac{t}{1.4 \times 10^{17}}=\frac{1}{0.7} \\ t\left(\frac{1}{0.23}-\frac{1}{1.4}\right)=\frac{10^{17}}{0.7} \\ t(3.64)=\frac{10^{17}}{0.7} \\ t=\frac{10^{17}}{0.7 \times 3.64}=\frac{10}{2.56} \times 10^{16} \\ =3.9 \times 10^{16}=4 \times 10^{16} \mathrm{~s}\end{aligned}$