Correct Answer  Option 1 : Bulb A
The correct answer is option 1) i.e. Bulb A
CONCEPT:

Power: The rate of work done by an electric current is called power. It is denoted by P. The SI unit of power is the watt (W).
Power dissipation is given by:
\(⇒ P = VI = \frac{V^2}{R}=I^2R\)
Where V is the potential difference across resistance, I is current flowing and R is resistance.
EXPLANATION:
Given that:
The power rating of bulb A, P_{A} = 50 W
The power rating of bulb B, PB = 100 W
We know, \(P =\frac{V^2}{R} ⇒ R =\frac{V^2}{P}\)
 The resistance of bulb A,
\(\Rightarrow R_A =\frac{V^2}{P_A} = \frac{200^2}{50} =800\: Ω\)
 The resistance of bulb B,
\(⇒ R_B =\frac{V^2}{P_B} = \frac{200^2}{100} =400\: Ω\)
 The equivalent resistance provided by the bulbs connected in series,
⇒ R = 800 + 400 = 1200 Ω
 The current in the circuit,
\(⇒ I =\frac{V}{R} = \frac{200}{1200} = 0.16\:A\)
The bulb which dissipates more power will glow brighter.
Power dissipated by the bulbs:
⇒ P_{A} = VI = I^{2}R_{A} = 0.16^{2} × 800 = 20.48 W
⇒ PB = VI = I2RB = 0.162 × 400 = 10.24 W
⇒ PA > PB
 Hence, bulb A will glow brighter.
 In a series connection, both bulbs will have the same current flowing through them.
 Therefore, the bulb with greater resistance will have a greater voltage drop across it and will have a higher power dissipation and brightness.