Power determination for resistive load
From your readings you can calculate the effective resistance of the load at \$ R = \frac {V}{I} = \frac {155}{3} = 52 \ \Omega \$.

If a resistance of this value was connected to a 230 V supply then the power would be given by \$ P = \frac {V^2}{R} = \frac {230^2}{52} = 1 \ \text {kW} \$.

You would need to check:


Loss of one or more heater elements. My bathroom heater has 2 x 1 kW elements in parallel and one is switched off by thermostat once the air reaches a certain temperature.
Power selector switch.
Any kind of dimmer power control.


I have assumed that the fan motor is a small load - maybe 50 W - in comparison with the heaters.