A radio isotope ‘ X ’ with a half life 1.4 × 10 ^ 9...

Given a radioactive isotope decays into a stable element,

$X \to Y$

At time $t = 0$, \left[x\right] = {N}_{0}$, \left[y\right] = 0$

At time $t = t$, \left[x\right] = {N}_{0} - x$, \left[y\right] = x$

$\dfrac{{N}_{0} - x}{x} = \dfrac{1}{7} ⇒ x = \dfrac{7{N}_{0}}{8}$
So, remaining nuclei of $X = {N}_{0} - x = \dfrac{{N}_{0}}{8} = \dfrac{{N}_{0}}{{2}^{3}}$
So, three half-life periods would have been passed,
$⇒ t = 3{t}_{\dfrac{1}{2}} = 3 \times 1.4 \times {10}^{9} = 4.2 \times {10}^{9}$ years.