Answer
$T=7.64h$
Work Step by Step
We can find the required orbital time period as
$T=(\frac{2\pi}{\sqrt{GM}})r^{\frac{3}{2}}$
We plug in the known values to obtain:
$T=\frac{2\pi}{\sqrt{6.67\times 10^{-11}(0.108\times 5.97\times 10^{24})}}(9378)^{\frac{3}{2}}$
$T=27500\ s$
$T=7.64\ h$
