Let $$N_{235}(0)$$ be the initial number of atoms of $${}^{235} \mathrm{U}$$
$$ N_{238}(0)$$ be the initial number of atoms of $${ }^{238} \mathrm{U},$$
$$ N_{235}(t)$$ be the number of atoms of $${ }^{235} \mathrm{U}$$ at any time $$t, $$
$$N_{238}(t)$$ be the number of atoms of $$^{238} \mathrm{U}$$ at any time $$t,$$
We knoe that the exponential decay expression can be written as:
$$N_{235}(t)=N_{5}(0) e^{-\lambda_{235} t}$$
$$N_{238}(t)=N_{8}(0) e^{-\lambda_{238} t}$$
On solving:
$$\dfrac{N_{235}(t)}{N_{238}(t)}=\dfrac{N_{235}(0)}{N_{238}(0)} e^{-\left(\lambda_{235}-\lambda_{238}\right) t}$$
solve for $$t$$ to get:
$$e^{-\left(\lambda_{235}-\lambda_{238}\right) t}=\left(\dfrac{N_{235}(t)}{N_{238}(t)}\right)\left(\dfrac{N_{238}(0)}{N_{235}(0)}\right)$$
$$\left(\lambda_{238}-\lambda_{235}\right) t=\ln \left(\left(\dfrac{N_{235}(t)}{N_{238}(t)}\right)\left(\dfrac{N_{238}(0)}{N_{235}(0)}\right)\right)$$
$$t=\left(\lambda_{238}-\lambda_{235}\right)^{-1} \ln \left(\left(\dfrac{N_{235}(t)}{N_{238}(t)}\right)\left(\dfrac{N_{238}(0)}{N_{235}(0)}\right)\right)$$
where
$$\lambda_{238}=\dfrac{\ln (2)}{T_{1 / 2,238}} \quad \lambda_{235}=\dfrac{\ln (2)}{T_{1 / 2,235}}$$
thus:
$$t=\left(\dfrac{\ln (2)}{T_{1 / 2,238}}-\dfrac{\ln (2)}{T_{1 / 2,235}}\right)^{-1} \ln \left(\left(\dfrac{N_{235}(t)}{N_{238}(t)}\right)\left(\dfrac{N_{238}(0)}{N_{235}(0)}\right)\right)$$
It is known that the half lives of $${ }^{235} \mathrm{U}$$ and $${ }^{238} \mathrm{U}$$ are $$7.04 \times 10^{8} \mathrm{y}$$ and $$44.7 \times 10^{8} \mathrm{y},$$ respectively,
substitute with these values $$N_{235}(t) / N_{238}(t)=0.0072$$ and $$N_{8}(0) / N_{5}(0)=0.15$$ to get:
$$t=\left(\dfrac{\ln (2)}{44.7 \times 10^{8} \mathrm{y}}-\dfrac{\ln (2)}{7.04 \times 10^{8} \mathrm{y}}\right)^{-1} \ln \left((0.0072)(0.15)^{-1}\right)$$
$$=3.66 \times 10^{9} \mathrm{y}$$
$$\quad t=3.66 \times 10^{9} \mathrm{y}$$
so time is $$3.66 \times 10^{9} \mathrm{year}$$