Let's determine the mass step by step:

The bullet is heated from 300 K to 600 K, so the increase in temperature is:

$$\Delta T = 600\, \text{K} - 300\, \text{K} = 300\, \text{K}.$$

The heat required to raise the temperature of the bullet (using the specific heat capacity of lead) is:

$$Q_1 = m \times c \times \Delta T = m \times 125\, \text{J/kg K} \times 300\, \text{K} = 37500\, m \, \text{J},$$

where $ m $ is the mass in kilograms.

After reaching 600 K, the bullet melts. The heat required for the phase change (melting) is given by:

$$Q_2 = m \times L = m \times 2.5 \times 10^4\, \text{J/kg} = 25000\, m \, \text{J}.$$

The total heat required for the process is:

$$Q = Q_1 + Q_2 = 37500\, m + 25000\, m = 62500\, m \, \text{J}.$$

We are given that the total heat is 625 J, so:

$$62500\, m = 625.$$

Solving for $ m $:

$$m = \frac{625}{62500} = 0.01\, \text{kg}.$$

Finally, converting the mass from kilograms to grams:

$$0.01\, \text{kg} = 0.01 \times 1000\, \text{g} = 10\, \text{g}.$$

Thus, the mass of the bullet is 10 grams. This corresponds to Option D.