Resistance of bulb, Rb=PbV2=60120×120=240Ω
Resistance of heater, Rh=PhV2=240120×120=60Ω

Voltage across bulb before the heater is not connected, V1=Rb+6V×Rb=246120×240V=117.07V
Now the bulb and heater are connected in parallel.
Resistance of the combination is Rnet=240+60240×60=48Ω
Voltage across bulb after the heater is switched on,
V2=Rnet+6V×Rnet=48+6120×48=106.66V
The decrease in the voltage is V1−V2=117.07−106.66≃10.4V
Note: Here supply voltage is taken as rated voltage.