Answer
$5\times10^{8}$ years.
Work Step by Step
Mass of $\,^{238}U$ decayed to $\,^{206}Pb$=
$0.08\,mg\,\,^{206}Pb\times\frac{238\,mg\,\,^{238}U}{206\,mg\,\,^{206}Pb}=0.0924\,mg$
Original mass of $^{238}U=$$1.09\,mg+0.0924\,mg$
$=1.1824\,mg$
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{4.51\times10^{9}\,y}=1.5366\times10^{-10}\,y^{-1}$
Recall that
$\ln(\frac{original\,mass}{present\,mass})=kt$
$\implies \ln(\frac{1.1824}{1.09})=0.08137=1.5366\times10^{-10}\,y^{-1}(t)$
$\implies t=\frac{0.08137}{1.5366\times10^{-10}y^{-1}}=5\times10^{8}\,y$
