B

Step 1: Calculate the resistance of each bulb using the formula for power:
\( P = \frac{V^2}{R} \)
For each bulb:
\( 500 = \frac{(220)^2}{R} \)
Thus,
\( R = \frac{(220)^2}{500} = \frac{48400}{500} = 96.8 \, \Omega \)

Step 2: In a series circuit, the total resistance is the sum of the individual resistances:
\( R_{total} = R_1 + R_2 = 96.8 + 96.8 = 193.6 \, \Omega \)

Step 3: Calculate the total current in the circuit using Ohm's Law:
\( V = I R \)
For the 110 volt line:
\( 110 = I \times 193.6 \)
Therefore,
\( I = \frac{110}{193.6} = 0.568 \, A \)

Step 4: Calculate the power generated by each bulb using the formula:
\( P = I^2 R \)
For one bulb:
\( P = (0.568)^2 \times 96.8 \)
which calculates to approximately 32.83 watts. Thus, the power generated by each bulb is 32.83 watts.

Therefore, the answer is 32 watts for each bulb, so option B.