Answer
$Q_{heater}=15.83kW$
Work Step by Step
To maintain the house at constant temperature the internal energy does not change. Then:
$E_{in}-E_{out}=0$
$E_{in}=E_{out}$
Since there is no work:
$Q_{in}=Q_{out}$
$Q_{people,lights and appliances}+Q_{heater}=Q_{out}$
$Q_{heater}=60000\frac{Btu}{h}-6000\frac{Btu}{h}=54000\frac{Btu}{h}$
$Q_{heater}=54000\frac{Btu}{h}*(\frac{1kW}{3412\frac{Btu}{h}})=15.83kW$
