Solution

The work done in pushing the cork is equal to the change in gravitional potential energy of the system. This is equal to the gravitational potential of the additional water at the surface with the same volume as the water replaced by the cork, which is at the zero-potential level. The volume $\pi r^2l$ of replaced water becomes a water disk of thickness $(\pi r^2l)/(\pi R^2)=r^2l/R^2$ at the surface, and its center of mass is at the height $h+r^2l/(2R^2)$ above the zero-potential level. Thus, its gravitational potential energy is

$$\rho\pi r^2lg\left(h+\frac{r^2l}{2R^2}\right).$$