MOSFET thermal resistance
You have calculated that your application circuit will dissipate 235W. To keep the junction below a reasonable temperature (I do not like to get closer to the maximum than perhaps 50C) lets set a maximum junction of 125C.

For your device, that means for an ambient condition of 25C and a maximum junction of 125C (100C temperature rise), you will need a sink that has a total thermal resistance of 0.4255C/W.

Unfortunately, the junction to case and case to sink parameters, when added together exceed this by a significant margin (0.53C/W).

The best in this situation would be to use multiple devices in parallel if possible.

The contribution of the above parameters alone will cause a temperature rise of 124.55C (for a junction temperature of 149.55C for 25C ambient), so the heat sink would need to have a thermal resistance of no more than 0.1C/W and even then the device will be running at the maximum permitted junction temperature which is highly not recommended.

Operating at absolute maximum ratings will definitely shorten the device life quite apart from being hot enough to be dangerous.

Although such heat sinks exist, they would have forced air and be very bulky, as you can see from this list and as you can see, they are quite expensive.

Note that this analysis does not take into account a heat sink that has imperfect fitting, so no margin is available.

So my suggestion is to go back to the design and find a method of achieving your requirements without such a large amount of heat in one place.

Maximum power handling

This is an area that is often misunderstood; the part you are using has a maximum power capability limited by the internals of the part, but the actual maximum power dissipation allowed is determined by the thermal resistance.

From this app note from Infineon there is a graph of maximum power handling at ambient with no heat sinks:


As you can see, the D2PAK maxes out at just over 5W for 25C ambient.