The correct option is A 12.22
The power across heater coil is
P=V2hRh
V2h=PRh
V2h=110×110
Vh=110V
Since resistance R and coil are in parallel voltage across them is same.
⇒ Voltage across 11Ω resistor is also 110 V.
Now, current across 11Ω resistor is
I=11011
I=10A
Also, current through coil is
Ih=VhRh
Ih=110110
Ih=1A
Same current equal to current through 11Ω resistor will flow through the parallel combination of coil and R. Hence, current through R is I−Ih=10−1=9A
Now, resistance R is
R=VRIR=1109=12.22Ω