Answer
$1.2\times10^{6}\,years$
Work Step by Step
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{2.44\times10^{5}\,y}=2.84\times10^{-6}\,y^{-1}$
Original amount $A_{0}=2.80\,kg$
Amount remaining $A=0.10\,kg$
Recall that $\ln(\frac{A_{0}}{A})=kt$ where $t$ is the time required.
$\implies \ln(\frac{2.80\,kg}{0.10\,kg})=3.3322=2.84\times10^{-6}\,y^{-1}(t)$
$\implies t=\frac{3.3322}{2.84\times10^{-6}\,y^{-1}}=1.2\times10^{6}\,y$
