Radioactive decay follows first order kinetics.  $\text{L}\text{e}\text{t}{[\text{A}]}_0=100$

 $\begin{array}{l}\therefore [\text{A}]=100\times 80\text{\%}=80\\ \text{D}\text{e}\text{c}\text{a}\text{y}\text{constant}\text{}(\text{k})=\frac{0.693}{{\text{t}}_{1/2}}=\frac{0.693}{5730}\\ \text{t}=\frac{2.303}{\text{k}}\text{log}\frac{[\text{A}\text{o}]}{[\text{A}]}\\ =\frac{2.303}{\frac{0.693}{5730}}\text{log}\frac{100}{80}\\ =\frac{2.303\times 5730}{0.693}\times \text{log}1.25\\ =\frac{2.303\times 5730}{0.693}\times 0.0969=1845\text{ yrs}\\ \end{array}$