Q.A light bulb is rated 100 W for a 220 V AC supply of 50 Hz. Calculate

[!TLDR] $R=V^2/P=220^2/100=484\,\Omega$; $i_{rms}=P/V=100/220=0.4545$ A. [!ANSWER] R = 484 $\Omega$; $i_{rms}=0.4545$ A.

The bulb is rated 100 W at 220 V (rms) AC, 50 Hz. Since a resistor (the bulb's filament) behaves identically for AC and DC, its power rating obeys the usual $P=\dfrac{V_{rms}^2}{R}$, so the resistance is $R=\dfrac{V_{rms}^2}{P}=\dfrac{220^2}{100}=\dfrac{48400}{100}=484\,\Omega$.\n\nThe rms current follows from $i_{rms}=\dfrac{V_{rms}}{R}=\dfrac{220}{484}\approx0.4545$ A (equivalently $i_{rms}=P/V_{rms}=100/220\approx0.4545$ A). [!ANSWER] (a) $R=484\,\Omega$.

(b) $i_{rms}=0.4545$ A.

Use the rms-value form of the power formula, $R=V_{rms}^2/P$, then Ohm's law with rms values for the current.

Using the rated wattage together with an assumed PEAK voltage instead of the given rms (220 V is already rms for an AC mains rating), which would give an incorrect resistance.