Answer
$$\frac{H}{R}=0.005$$
Work Step by Step
The mass of earth is called $M_e$ and the radius of earth is $R$.
The weight of the probe with mass $m$ on the surface of earth is $$W_{sur}=G\frac{mM_e}{R^2}$$
The weight of the same probe at a distance $H$ above the surface of earth is $$W_{H}=G\frac{mM_e}{(R+H)^2}$$
From the given information, we have $$W_H=0.99W_{sur}$$ $$G\frac{mM_e}{(R+H)^2}=0.99G\frac{mM_e}{R^2}$$ $$\frac{R^2}{(R+H)^2}=0.99$$ $$\frac{R}{R+H}=0.995$$ $$0.005R=0.995H$$ $$\frac{H}{R}=\frac{0.005}{0.995}\approx0.005$$
