Answer
$ 8.67 \times10^7\;\rm m$
Work Step by Step
First, we need to find the periodic time of the satellite that will rotate around the moon above the same position. This means that the periodic time of the satellite is the same periodic time as the moon's full rotation around its own axis.
We know that the moon completes one full rotation around its axis every 27.3 days.
Thus, $$T_{S}=\bf 27.3 \bf \;\rm day$$
The centripetal acceleration needed for the satellite's circular motion around the moon is applied by the gravitational force between the moon and the satellite.
Hence,
$$\sum F_S=F_G=m_{S}a_r$$
$$\dfrac{Gm_M \color{red}{\bf\not}m_S}{ r^{ \color{red}{\bf\not}2}}= \color{red}{\bf\not}m_s\dfrac{v^2}{ \color{red}{\bf\not}r}$$
where $r$ is the distance between the center of the moon and the satellite which is given by $r=R_M+h$
$$\dfrac{Gm_M }{ R_M+h}= v^2 $$
Noting that the speed of the satellite is given by $v=2\pi r/T=2\pi(R_M+h)/T$
$$\dfrac{Gm_M }{ R_M+h}= \left[\dfrac{2\pi (R_M+h)}{T} \right]^2$$
Hence,
$$\dfrac{Gm_M }{ R_M+h}= \dfrac{4\pi^2 (R_M+h)^2}{T^2} $$
$$(R_M+h)^3=\dfrac{Gm_MT^2}{4\pi^2}$$
$$(R_M+h) =\sqrt[3]{\dfrac{Gm_MT^2}{4\pi^2}}$$
$$h =\sqrt[3]{\dfrac{Gm_MT^2}{4\pi^2}}-R_M$$
Plugging the known;
$$h =\sqrt[3]{\dfrac{(6.67\times 10^{-11})(7.36\times 10^{22})(27.3\times 24\times 60^2)^2}{4\pi^2}}-(1.74\times 10^6)$$
$$h=\color{red}{\bf 8.67 \times10^7}\;\rm m$$
