Calculating Wattage for Gate Resistor for IRFP460 MOSFET used in High Frequency [400KHz] Application?
Dividing the peak power by the frequency is not useful.

Instead, you would multiply it by the duty cycle. If you're dumping 25 W of power into the resistor for 2 &times; 100 ns out of every 2.5 &micro;s. This would be an average power of

$$25 W \cdot\frac{2 \cdot 100 ns}{2.5 \mu s} = 2 W$$

Clearly, your 1W resistor is not going to cut it!

However, the peak instantaneous power is not really a good estimate of the average power during the switching transient. A better estimate can be arrived at by considering the energy flow into and out of the gate capacitance.

For an R-C circuit, the energy dissipated in the resistor is basically equal to the energy that ends up on the capacitor. If your gate charge is 210 nC and your gate voltage is 12V, this represents

$$Energy = \frac{1}{2}\cdot Charge \cdot Voltage$$

$$0.5 \cdot 210 nC \cdot 12 V = 1.26 \mu J$$

This is the energy you're dumping into the gate capacitance, and then dumping out again on every switching cycle. All of this energy gets dissipated in the gate resistor.

To get the average power, multiply the energy per cycle by the number of cycles per second, giving

$$1.26 \mu J \cdot 2 \cdot 400 kHz = 1.088 W$$

Your 1W resistor would be running at its limit, with no margin. I would use a 2W resistor here.