Answer
280 years.
Work Step by Step
Amount of radionuclide at the beginning $A_{0}=100$
Amount of radionuclide remaining
$A= 100.0-99.9=0.1$
Rate constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{28\,y}=0.02475\,y^{-1}$
$\ln(\frac{A_{0}}{A})=kt$
$\implies \ln(\frac{100}{0.1})=6.9=0.02475\,y^{-1} \times t$
Or $t= \frac{6.9}{0.02475\,y^{-1} }=280\,y$
