Answer
Again, we use energy conservation.
$$
K_{i}+U_{i}=K_{f}+U_{f} \Rightarrow \frac{1}{2} m\left(3.70 \times 10^{3}\right)^{2}-\frac{G m M}{R_{0}}=0-\frac{G m M}{R_{f}}
$$
Therefore, we find $$R_{f}=7.40 \times 10^{6} \mathrm{m} .$$ This corresponds to a distance of 1034.9 $$\mathrm{km} \approx 1.03 \times 10^{3} \mathrm{km}$$ above the Earth's surface.
Work Step by Step
Again, we use energy conservation.
$$
K_{i}+U_{i}=K_{f}+U_{f} \Rightarrow \frac{1}{2} m\left(3.70 \times 10^{3}\right)^{2}-\frac{G m M}{R_{0}}=0-\frac{G m M}{R_{f}}
$$
Therefore, we find $$R_{f}=7.40 \times 10^{6} \mathrm{m} .$$ This corresponds to a distance of 1034.9 $$\mathrm{km} \approx 1.03 \times 10^{3} \mathrm{km}$$ above the Earth's surface.
