The diameter is 10^-6 what? meters? That is what I will assume.
Assume each micrometeorite occupies a volume of (10^-6 m)^3 = 10^-18 m That is close enough, since they will not be the same size and not perfectly packed in a hexagonal array. About half the space will be void.
Divide the (1.5)^3 m^3 = 3.38 m^3 box by 10^-18 for the number of particles needed to fill the box. It will take that many seconds to fill. My calculated number exceeds the age of the moon, and the universe
If one micrometeorite (a sphere with a diameter of 1.0 multiplied by 10-6) struck each square meter of the moon each second, it would take many years to cover the moon with micrometeorites to a depth of 1.5 m. Consider a cubic box, 1.5 m on a side, on the moon. Estimate how long it would take to completely fill the box with micrometeorites if one micrometeorite landed in the box each second.
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