Answer
8.21x$10^{-6}$
Work Step by Step
The Moon's spin angular momentum can be calculated by $$L_{spin}=I_{spin}\omega_{spin}=\frac{2}{5}MR^2_{Moon}\omega_{spin}$$
The orbital angular momentum can be calculated by $$L_{orbit}=I_{orbit}\omega_{orbit}=MR^2_{orbit}\omega_{orbit}$$
Because the same side of the Moon always face the Earth, $\omega_{spin}=\omega_{orbit}$.
$$\frac{L_{spin}}{L_{orbit}}=\frac{\frac{2}{5}MR^2_{Moon}\omega_{spin}}{MR^2_{orbit}\omega_{orbit}}=\frac{2}{5}(\frac{R_{Moon}}{R_{orbit}})^2=0.4(\frac{1.74\times10^6m}{3.84\times10^8m})^2$$
$$=8.21\times10^{-6}$$
