Answer
0.35 lunar month.
Work Step by Step
Let T be the period of revolution of the moon and T' be the period of revolution of the satellite. Then
From Kepler's law of periods, we have
$T^{2}=(\frac{4\pi^{2}}{GM})r^{3}$ and
$(T')^{2}=(\frac{4\pi^{2}}{GM})(r')^{3}$
$\implies \frac{(T')^{2}}{T^{2}}=(\frac{r'}{r})^{3}$
Or $T'=\sqrt {(\frac{r'}{r})^{3}\times T^{2}}$
Given that $T= 1\,lunar\,month$ and $r'=\frac{1}{2}r$
Therefore, we get
$T'=\sqrt {(\frac{1}{2})^{3}\times(lunar\,month)^{2}}=0.35\,lunar\,month$
