Answer
The inverse-square nature of gravity tells us that if we are twice as far from the Earth’s center, the gravitational field will be $\frac {1}{4}$ its value at the surface, or $2.5 \frac{m}{s^{2}}$.
$$g = G \frac{m_{Earth}}{d^{2}} = (6.67 \times 10^{-11} \frac {N \cdot m^{2}}{kg^{2}}) \frac{6.0 \times 10^{24} kg}{(1.28 \times 10^{7} m)^{2}} = 2.5 \frac{m}{s^{2}} $$
