1. Use the bulb's rated values to find its resistance

For a bulb rated $50\,\text{W}$ at $200\,\text{V}$,

$P = \frac{V^2}{R}$

So,

$R = \frac{V^2}{P} = \frac{(200)^2}{50} = \frac{40000}{50} = 800\,\Omega$

2. Now connect it to the actual supply voltage

The bulb is connected across $100\,\text{V}$, so the power dissipated becomes

$P' = \frac{V'^2}{R} = \frac{(100)^2}{800} = \frac{10000}{800} = 12.5\,\text{W}$

3. Match with the options

$12.5\,\text{W}$

So the correct option is C.

4. Verification with stored answer

Stored correct answer: C

My derived answer: C

They agree.