Here, $ m = 1800 \,kg $
Frictional force, $ f= 4000\, N $
Uniform speed, $ v = 2\, ms^{-1} $
Downward force on elevator is
$ F = mg+f= (1800 \,kg \times 10\, ms^{-2}) + 4000\, N = 22000 \,N $
The motor must supply enough power to balance this force. Hence,
$ P = Fv = (22000 \,N)(2\,ms^{-1}) $
$ = 44000 \,W = 44 \times 10^3\, W = 44\, kW $