Answer
$ 0.414R $
Work Step by Step
According to Newton's gravitational law, the free-fall acceleration at the planet's surface is given by
$$g=\dfrac{GM}{R^2}\tag 1$$
where $M$ is the planet's mass and $R$ is its radius.
The author told us that at some height $h$, the astronaut measured the free-fall acceleration and found it is half the free-fall acceleration at the ground.
Thus,
$$g'=\dfrac{GM}{(R+h)^2}=\frac{1}{2}g$$
Plugging from (1);
$$\dfrac{ \color{red}{\bf\not}G \color{red}{\bf\not}M}{(R+h)^2}=\dfrac{ \color{red}{\bf\not}G \color{red}{\bf\not}M}{2R^2}$$
Hence,
$$h+R=\sqrt2 R$$
So, their altitude is
$$\boxed{h=(\sqrt2 -1)R=\color{red}{\bf0.414}R}$$
