At the time of observation (t = t),

      $\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}=\frac{100}{1}$            (given)

Further, it is given that $\frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}=\frac{1.02}{1}$

Number of atoms $\mathrm{N}=\frac{\mathrm{m}}{\mathrm{A}}$

$\therefore \frac{{\mathrm{N}}_1}{{\mathrm{N}}_2}=\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}\times \frac{{\mathrm{A}}_2}{{\mathrm{A}}_1}=\frac{100}{1.02}$   …….(i)

Let N₀ be the number of atoms of both the isotopes at the time of formation, then

             $\frac{{\mathrm{N}}_1}{{\mathrm{N}}_2}=\frac{{\mathrm{N}}_0{\mathrm{e}}^{-{\mathrm{\lambda }}_1\mathrm{t}}}{{\mathrm{N}}_0{\mathrm{e}}^{-{\mathrm{\lambda }}_2\mathrm{t}}}={\mathrm{e}}^{\left({\mathrm{\lambda }}_2-{\mathrm{\lambda }}_1\right)\mathrm{t}}$  ………(ii)

Equating Eqs. (i) and (i), we have

           ${\mathrm{e}}^{\left({\mathrm{\lambda }}_2-{\mathrm{\lambda }}_1\right)\mathrm{t}}=\frac{100}{1.02}$

$\begin{array}{l}\text{ or }\left({\mathrm{\lambda }}_2-{\mathrm{\lambda }}_1\right)\mathrm{t}=\ln \left(100\right)-\ln (1.02)\\ \therefore \mathrm{t}=\frac{\ln \left(100\right)-\ln (1.02)}{\left(\frac{1}{2\times {10}^9}-\frac{1}{4\times {10}^9}\right)}\end{array}$

Substituting the values, we have

               $\mathrm{t}=1.834\times {10}^{10}\mathrm{yr}$