1. Given data

• Mass of ice: $m_i = 120\,\text{g} = 0.12\,\text{kg}$ at $0^\circ\text{C}$
• Mass of water: $m_w = 300\,\text{g} = 0.30\,\text{kg}$ at $25^\circ\text{C}$
• Final temperature: $0^\circ\text{C}$
• Specific heat capacity of water: $c = 4200\,\text{J kg}^{-1}\text{K}^{-1}$
• Latent heat of fusion of ice: $L = 3.5 \times 10^5\,\text{J kg}^{-1}$

2. Heat lost by water in cooling from $25^\circ\text{C}$ to $0^\circ\text{C}$

$Q_{\text{lost}} = m_w c \Delta T$

$Q_{\text{lost}} = 0.30 \times 4200 \times 25$

$Q_{\text{lost}} = 31500\,\text{J}$

3. Heat required to melt $x$ g of ice

If $x$ g of ice melts, then melted mass in kg is

$\frac{x}{1000}$

Heat required for melting:

$Q_{\text{gain}} = \frac{x}{1000} \times 3.5 \times 10^5$

4. Apply heat balance

Since final temperature is $0^\circ\text{C}$, all heat lost by water is used only to melt some ice:

$31500 = \frac{x}{1000} \times 3.5 \times 10^5$

$31500 = 350x$

$x = \frac{31500}{350} = 90$

5. Check

Initial ice mass is $120\,\text{g}$, so melting of $90\,\text{g}$ is possible.

Hence, the amount of ice melted is:

$\boxed{90}$