Answer
$1.1\times10^{4}\,y$
Work Step by Step
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730\,y}=1.2094\times10^{-4}\,y^{-1}$
Original amount $A_{0}=100$
Amount remaining $A=25$
Recall that $\ln(\frac{A_{0}}{A})=kt$ where $t$ is the time required.
$\implies \ln(\frac{100}{25})=1.3863=1.2094\times10^{-4}\,y^{-1}(t)$
$\implies t=\frac{1.3863}{1.2094\times10^{-4}\,y^{-1}}=1.1\times10^{4}\,y$
