Power of bulb $$=60W$$ $$\left( {given} \right)$$

Resistance of bulb $$ = {{120 \times 120} \over {60}} = 240\Omega $$

$$\left[ {\,\,} \right.$$ $$\left. {\,P = {{{V^2}} \over R}\,} \right]$$

Power of heater $$=240W$$ (given)

Resistance of heater $$ = {{120 \times 120} \over {240}} = 60\Omega $$

Voltage across bulb before heater is switched on,

$${V_1} = {{240} \over {246}} \times 120 = 117.73\,\,$$ volt

Voltage across bulb after heater is switched on,

$${V_2} = {{48} \over {54}} \times 120 = 106.66$$ volt

Hence decrease in voltage

$${V_1} - {V_2} = 117.073 - 106.66 = 10.04$$ Volt (approximately)