Okay, so my very first number was approximately right. 
I don’t know if you’ve noticed, Chronos, but it’s not a trivial amount of matter you’re kicking up - it’s about one-third the mass of the moon. And the energy you’re putting in is about half the total binding energy of the moon. I think that this will affect the answer - because as the mass of the moon decreases so will its escape speed - but probably not by an order of magnitude.
Okay, I worked it out. If we kick up a bit of mass dm at a time, and we assume that the moon maintains a uniform density so that r is proportional to m[sup]1/3[/sup], then v[sub]esc[/sub] = am[sup]1/3[/sup], where a = m[sub]0[/sub][sup]1/6/sup[sup]1/2[/sup] = 5.671e-5 [mks]. The momentum of the particle is v[sub]esc[/sub]dm, and so the rest of the moon slows by an amount dv = v[sub]esc[/sub]dm/m = am[sup]-2/3[/sup]dm. Integrating gives Delta-v = 3a(m[sub]1[/sub][sup]1/3[/sup] - m[sub]0[/sub][sup]1/3[/sup]) = 3Delta-v[sub]esc[/sub]. Meanwhile, the energy spent is dE = v[sub]esc[/sub][sup]2[/sup]/2 dm, and similar integration gives E = 3/10 Delta(mv[sub]esc[/sub][sup]2[/sup]).
For the problem in this thread, Delta-v = 812.7 m/s, from which you can get v[sub]esc,final[/sub] and so on. m[sub]final[/sub] = 5.111e22 kg, and E = 5.649e28 J. So, you’d wind up with 70% of the moon, but you’d get it to Earth, darn it!
Would it have to be a nuclear bomb? Couldn’t it be a meteor? And, is an impact absolutely necessary? Couldn’t someon slingshot a large asteroid around the moon so that the asteroid takes energy away from the moon?
Wouldn’t a 2020’s-Style Death Ray be a lot more efficient than a 1920’s -Style?
Just in case you’re worried about the moon crashing into Earth, it’s actually flying away into space.