Let the half-lives of $U^{235}$ and $U^{238}$ be $T_1$ and $T_2$ respectively, and their decay constants be $\lambda_1$ and $\lambda_2$.

Given,

$T_1 = 7.13 \times 10^8 \text{ years}$

$T_2 = 4.5 \times 10^9 \text{ years}$

The decay constants are calculated as follows:

$$ \begin{aligned} \lambda_1 &= \frac{0.693}{T_1} = \frac{0.693}{7.13 \times 10^8} \approx 9.72 \times 10^{-10} \text{ year}^{-1} \\ \lambda_2 &= \frac{0.693}{T_2} = \frac{0.693}{4.5 \times 10^9} \approx 1.54 \times 10^{-10} \text{ year}^{-1} \end{aligned} $$

According to the problem, at the beginning of the Earth's formation, the amounts of $U^{235}$ and $U^{238}$ were equal, i.e., $N_{01} = N_{02} = N_0$. Currently, the ratio of their amounts is $N_1 : N_2 = 1 : 140$. Thus, $\frac{N_2}{N_1} = 140$.

Using the radioactive decay law $N = N_0 e^{-\lambda t}$ for both isotopes:

$$ \begin{aligned} N_1 &= N_0 e^{-\lambda_1 t} \quad \dots (i) \\ N_2 &= N_0 e^{-\lambda_2 t} \quad \dots (ii) \end{aligned} $$

Dividing equation (ii) by equation (i):

$$ \begin{aligned} \frac{N_2}{N_1} &= \frac{N_0 e^{-\lambda_2 t}}{N_0 e^{-\lambda_1 t}} \\ 140 &= e^{(\lambda_1 - \lambda_2)t} \\ \ln(140) &= (\lambda_1 - \lambda_2)t \\ t &= \frac{\ln(140)}{\lambda_1 - \lambda_2} \end{aligned} $$

Substituting the values:

$$ \begin{aligned} t &= \frac{\ln(140)}{(9.72 \times 10^{-10}) - (1.54 \times 10^{-10})} \\ t &= \frac{4.9416}{8.18 \times 10^{-10}} \\ t &\approx 6.04 \times 10^9 \text{ years} \end{aligned} $$

Therefore, the age of the Earth is approximately $6.04 \times 10^9$ years.