On a large tray of mass M an ice cube of mass m and of edge L is kept. If the ice melts completely, the centre of mass of the system come down by

The height of water when the ice melts is little due to the size of the tray.
Shift in COM is essentially the level of COM before to melting ${\mathrm{y}}_{\mathrm{cm}}$.

$$\begin{gathered} =\frac{\mathrm{m}\times {\displaystyle \frac{\mathrm{L}}{2}}}{\mathrm{M}+\mathrm{m}} \\ =\frac{\mathrm{mL}}{2(\mathrm{M}+\mathrm{m})} \end{gathered}$$

Hence the correct answer is $\frac{\mathrm{mL}}{2(\mathrm{M}+\mathrm{m})}.$