Dmitryl, do you mean a load connected directly to the battery, or a load being supplied by the controller?
It should be possible to use a 3kW electric fire to load the battery directly, the current will vary according to the ratio between the fire's voltage and the battery's voltage.
3kW/240V = 12.5 Amps
240V/12.5A = 19.2 Ohms
48V x 19.2 Ohms = 2.5 Amps
48V x 2.5A = 120 Watts
So, if my theory is correct, a 240V 3kW resistive load should produce a 120W load on your 48V battery pack.
All you now need is eight electric fires
(connected together in parallel) to produce a 960W load.
Alan