The half-life for radioactive decay of

Here, $k=\frac{0.693}{t_{1 / 2}}$

$=\frac{0.693}{5730}$ years $^{-1}$

It is known that,

$t=\frac{2.303}{k} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$

$=\frac{2.303}{\frac{0.693}{5730}} \log \frac{100}{80}$

= 1845 years (approximately)

Hence, the age of the sample is 1845 years.