A well-insulated vessel, whose own heat capacity can be ignored, holds 0.100 kg of ice and 0.300 kg…

Step 1. The first ball (a). Cooling to $0\ ^\circ\text{C}$, it gives out $m c \Delta T = 0.200 \times 460 \times 300\ \text{J} = 27600\ \text{J}$. Melting all the ice would need $0.100 \times 3.36 \times 10^5\ \text{J} = 33600\ \text{J}$, which is more. So only part of the ice melts:

and everything is at $0\ ^\circ\text{C}$, with $17.9\ \text{g}$ of ice left.

Step 2. The second ball (b). Melting the rest of the ice needs $33600 - 27600 = 6000\ \text{J}$. After that, $0.400\ \text{kg}$ of water (heat capacity $1680\ \text{J/K}$) and the first ball ($92\ \text{J/K}$) warm to the final temperature $T$:

Step 3. More balls (c). Since energy is just shared out, the result after $n$ balls is the same as dropping all $n$ in together:

This gives $48.8\ ^\circ\text{C}$ for $n = 5$ and $59.1\ ^\circ\text{C}$ for $n = 6$. So the water first passes $50\ ^\circ\text{C}$ with the ball number $6$.