Resistance of bulb $=\frac{120 \times 120}{60}=240\, \Omega\left[\because \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}\right]$

Power of heater $=240\, \mathrm{W}$ (given)

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

Voltage across bulb before heater is switched on,

$\mathrm{V}_{1}=\frac{240}{246} \times 120=117.73\, \text { volt }$

Voltage across bulb after heater is switched on,

$\mathrm{V}_{2}=\frac{48}{54} \times 120=106.66\, \mathrm{volt}$

Hence decrease in voltage

$\mathrm{V}_{1}-\mathrm{V}_{2}=117.073-106.66=10.04\, \mathrm{Volt}$ (approximately)