A sample of rock contains 5.00 mg of uranium-238 and 0.90 mg of lead-206…

Step 1. The moles. Uranium: $\dfrac{5.00 \times 10^{-3}\ \text{g}}{238\ \text{g/mol}} = 2.101 \times 10^{-5}$ mol. Lead: $\dfrac{0.90 \times 10^{-3}\ \text{g}}{206\ \text{g/mol}} = 4.369 \times 10^{-6}$ mol.

Step 2. The ratio. $\dfrac{N_{Pb}}{N_U} = \dfrac{4.369 \times 10^{-6}}{2.101 \times 10^{-5}} = 0.208$.

Step 3. The age. $N_U = N_0e^{-\lambda t}$ and $N_U + N_{Pb} = N_0$. So $e^{\lambda t} = 1 + \dfrac{N_{Pb}}{N_U} = 1.208$. Then $t = \dfrac{T}{\ln 2}\,\ln 1.208 = \dfrac{4.47 \times 10^{9}\ \text{y}}{0.693} \times 0.189 = 1.22 \times 10^{9}$ y.