(N/A) Given: Rated voltage $V = 220 \, V$,Rated power $P = 60 \, W$.
$(i)$ Using the formula $P = \frac{V^2}{R}$,the resistance $R$ of the bulb is:
$R = \frac{V^2}{P} = \frac{(220)^2}{60} = \frac{48400}{60} \approx 806.67 \, \Omega$.
$(ii)$ When two identical bulbs are connected in series,the total resistance $R_T = R + R = 2R = 2 \times 806.67 = 1613.34 \, \Omega$.
The current $I$ flowing through the circuit is $I = \frac{V}{R_T} = \frac{220}{1613.34} \approx 0.136 \, A$.
The rate of energy conversion (power consumed) in each bulb is $P' = I^2 R = (0.136)^2 \times 806.67 \approx 15 \, W$.
$(iii)$ Total power consumed in series connection is $P_{total} = I^2 R_T = (0.136)^2 \times 1613.34 \approx 30 \, W$.
$(iv)$ If the bulbs are connected in parallel,the total power is $P_{total} = P_1 + P_2 = 60 + 60 = 120 \, W$.