Why we would know the angular momentum and need to solve the I rather than the other way around... Anyway, we can solve it as an exercise.
I may not be understanding the problem fully, but it seems like you can calculate time in the air (the usual calculation for leaving the ground with a velocity and getting to a given height:
t in the air = 2sqrt(2h/g) which is double the time it takes to drop from a height, h (rearranging d = 1/2 gt2)
I get .808 seconds. (a pretty decent hang time).
Given that we assume that this is a constant rotation process once he leaves the ground, you can solve for the angular velocity ω = 2πf f is the turns per second (frequency)
In this case ω = 2π(4.5 revs/.808 s) = 34.99 rads/ sec
We are given L, the angular momentum, and want I, the moment of inertia:
L = Iω or I = L/ω = 70 kg-m2/s / 34.99 rad/s = 2.0 kg-m2
Hope that helps.