Answer
$$F= G \frac{m_{1} m_{2}}{d^{2}} = (6.67 \times 10^{-11} \frac {N \cdot m^{2}}{kg^{2}}) \frac{(1 kg)(6.0 \times 10^{24} kg)}{(1.28 \times 10^{7} m)^{2}} = 2.4 N $$
The result is not surprising - by the inverse-square law, it is one-quarter the weight of a 1-kg mass at the surface.
