Answer
$$F= G \frac{m_{1} m_{2}}{d^{2}} = (6.67 \times 10^{-11} \frac {N \cdot m^{2}}{kg^{2}}) \frac{(1 kg)(6.0 \times 10^{24} kg)}{(6.4 \times 10^{6} m)^{2}} = 9.8 N $$
The result is not surprising - it is the weight of a 1-kg mass at the surface of the Earth.
