Short answer: "All else being equal", it looks to me like making Bizarro-Moon twice as big and three times as far away as the actual Moon can't happen -- not even if you spin the Earth down completely (ie. to the point where the Earth is tidally locked to the Moon, the way the Moon is already locked to the Earth).
So all else can't be equal. But there's a lot of things you could change, and you can get just about any answer you want. So 24 hour days is just fine.
Long answer: Bizarro-Moon is about twice as big and three times as far awaya, so we're talking three or four times the orbit angular momentum (which is proportional to mass and to the square-root of distancec).
The actual Earth lost about three quarters of its spin angular momentum (which is inversely proportional to spin periodd): days went from 6 hoursb to 24 hours. And the Moon also lost its own spin completely and is tidally locked -- but the Moon is a lot smaller than the Earth (angular momentum proportional to massd), so the Earth's spin angular momentum is a lot bigger. Bizarro-Moon is somewhat bigger but still a lot smaller than the Earth.
So: most of the Moon's orbit angular momentum came from the Earth's spin, most of that spin is gone now, and the result only got the actual-sized Moon to its actual position. You need three or four times more of that to get Bizarro-Moon where you want it -- so it won't get there, even if the Earth becomes completely tidally locked.
But maybe Bizarro-Earth is somewhat bigger than the actual Earth, and maybe the collision that produced Bizarro-Moon was imparted faster initial spin to Bizarro-Earth and/or a greater initial orbital angular momentum to Bizarro-Moon.
There's also the problem that Bizarro-Moon might be orbiting so far from Earth that its orbit is unstable due to Jupiter perturbing it over timee. But again, maybe Bizarro-Earth is somewhat bigger, so objects can have larger stable orbits.
You also talked about rate of disintegrationf, which I take to mean how much slower the Earth spins each year.
"All else being equal", if Bizarro-Moon is twice as massive, the rate of slowdown will be four times as fast (quadratic relation). If it's three times as far away, the rate of spin slowdown will be hundreds times slower (negative sixth power)g.
There's also an effect from the internal composition of Bizarro-Earth. 921
Maybe Bizarro-Earth has a different composition. Note that the internal composition of Bizarro-Moon won't matter to the Earth's spinh, but it would have mattered to how fast Bizarro-Moon lost its own spin (eg. how big are Ganymede's oceans reallyi?)
The dominant effect for the current rate is the fact that it is three times further away now, so transfer of momentum is going to be much much slower today. But, in the past, when Bizarro-Moon was closer to Bizarro-Earth, the other things matter.
Notes:
a Using OP's figures... After writing all of this, I later realized this wasn't clear. Just looking at OP's figures for diameter, you might conclude Bizarro-Moon about three times more massive than the actual Moon, not two. (about 50% bigger diameter than the actual Moon, and mass is proportional to diameter cubed). But I was also assuming that Bizarro-Moon has a similar density to Ganymede, which (not stated by OP) is about 2/3 the density of the Moon.
b Again using OP's figures, for the impact event -- I am not endorsing the giant impact hypothesis.
c Using the basic definition of angular momentum and Kepler's 3rd law: $ L = mr^2\omega $ and $ \omega^2r^3 = GM $.
d For spin angular momentum, $ L = \alpha mr^2\omega $, which is the same as orbit angular momentum, other than the fudge factor $\alpha$, which will be somewhat less than 0.4 (exactly 2/5 for a sphere of uniform density), but not much less, for a rocky moon or planet (it can be quite small for a gas giant or star). See moment of inertia factor.
e Basically, outside 1.5 million km, you are orbiting the Sun not the Earth. OP's figure for Bizarro-Moon is still inside this sphere. 921