In an elastic collision involving two objects both momentum and energy are conserved. You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore one object will have the opposite momentum of the other. Since energy is the square of the momentum divided by twice the mass, this means that the total kinetic energy is proportional the square of the momentum and this must thus be the same before and after the collision. the magnitude of the momentum therefore cannot change, we can thus conclude that all that can happen is that the momentum changes direction. In a one dimensional setting like in this problem, this means that the momenta of both object just changes sign.
So, using both energy and momentum conservation we've reduced all one dimensional collision problems to a mere triviality: Transform to the center of mass frame, change the sign of the momenta and then transform back to the original frame. Then in the case that one of the masses happens to be much larger than the other mass, the center of mass frame is the rest frame of that mass which makes it even easier to solve the problem.
Let's see if we can tackle this problem using the method I derived above. Here you need to consider that when the ball collides with the ground that's a collision of the big ball with the Earth obviously you should then take the big ball to the light mass. Also when that collision happens the small ball will initially keep on moving toward the ground and just a fraction later will it feel the impact due to the big ball having changed direction.
So, if the ball hits the ground with velocity v then this velocity will reverse sign as we're working in the rest frame of the Earth which is the relevant center of mass frame for that collision with the ground. But then what happens is that the small ball which is still moving at velocity v toward the ground hits the big ball that is moving upward with velocity v. In the rest frame of the big ball, the small ball is thus moving toward it with velocity 2 v. Since this frame is to a good approximation the relevant center of mass frame for the collision that is about to happen, it follows that after the collision the velocity will be 2 v relative to the big ball. Then transforming back the the original frame, we see that after the collision, the small ball will have a velocity of 3 v relative to the ground.
This means that the kinetic energy of the small ball after the collision will be 9 times what is was before the collision and that then implies that it will reach 9 times the height from which it was dropped.
Note: The outcome does not depend on the assumption that there be a small gap between the two balls
Some commentators and some other answers wrongly claim that the answer given here is wrong, because the Hyperphysics website claims that without there being a small gap between the balls, the outcome will be different. Now, apart from the fact the Hyperphysics websites make no such claim at all, the apparent dependence on the gap is an artifact of the way one treats the simultaneous collisions. Of course, one may argue that things are really a bit different when the collision happen simultaneously, but all the assumptions made should be made explicit, which they didn't do.
Simply put, if with a gap of zero the small ball doesn't rise as high, then what happened to the energy that would have gone into the small ball if the gap wasn't there? So, the hidden assumptions made leads to energy being dissipated in the form of vibrations in he the big ball. But as I show below, this is not at all obvious, you are not led to this conclusion if you just analyze the problem in more detail.
So, let me explain here why the gap doesn't matter in the spirit of this problem where we make the assumption of elastic collisions, where the two balls are certainly not glued to each other. The moment the bottom ball collides with the ground, the top ball together with the contact point are moving toward the ground with velocity v while the bottom part of the ball has come to a stop. The compression of the ball at the contact point at the ground thus happens first, the shock wave needs to propagate to the other end of the ball before the contact point with the other ball will come to a stop and the top ball will start to compress it there. When that happens the contact point with the top ball will end up being moved into the ball with the relative velocity of 2 v. This means that when that elastic motion rebounds this relative velocity of 2 v will reverse sign. After that point the contact point will slow down, creating a gap between the top ball (it's not glued to the bottom ball so it will move away). So, we again arrive at the same result.
Of course, one may argue that a more detailed treatment of the elastic motions of the balls is necessary and that the 2 v relative velocity isn't accurate. One has to note here that internal motion of the elastic balls is the process by which kinetic energy gets dissipated into heat, so this actually points to the impossibility of having a perfect elastic collisions to begin with.