Given: $\frac{\mathrm{d} \mathrm{N}_{0}}{\mathrm{dt}}=20$ decays $/ \mathrm{mir}$

$\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=2$ decays/min

$\mathrm{T}_{1 / 2}=5730 \mathrm{years}$

As we know,

$\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}$

$\log \frac{N_{0}}{N}=\lambda t$

$\therefore \mathrm{t}=\frac{1}{\lambda} \log \frac{\mathrm{N}_{0}}{\mathrm{N}}$

$ = \frac{{2.303 \times {{\text{T}}_{1/2}}}}{{0.693}} \times {\text{Lo}}{{\text{g}}_{10}}\frac{{{{\text{N}}_0}}}{{\text{N}}}$

But $\frac{\frac{d N_{0}}{d t}}{\frac{d N}{d t}}=\frac{N_{0}}{N}=\frac{20}{2}=10$

$\therefore \,\,t = \frac{{2.303 \times 5730}}{{0.693}} \times 1$

$=19039$ years