Calculate the energy consumed by the bulb in one day.

The power of the bulb is $P = 200 \text{ W}$.
The time it is lighted for is $t = 1 \text{ day} = 24 \text{ hours}$.
The energy consumed is given by the formula $E = P \times t$.

$$E = 200 \text{ W} \times 24 \text{ h}$$ $$E = 4800 \text{ Wh}$$

Since $1 \text{ kWh} = 1000 \text{ Wh}$, we divide the energy in Wh by 1000.

$$E = \frac{4800 \text{ Wh}}{1000 \text{ Wh/kWh}}$$ $$E = 4.8 \text{ kWh}$$

The cost of electricity is N 40 per kWh.
The total cost is the energy consumed (in kWh) multiplied by the rate per kWh.

$$\text{Cost} = E \times \text{Rate}$$ $$\text{Cost} = 4.8 \text{ kWh} \times \text{N } 40/\text{kWh}$$ $$\text{Cost} = \text{N } 192.00$$

The correct option is D.

The final answer is $\boxed{\text{N } 192.00}$.