When the current in an electric bulb decreases by 1%, the power decreases approximately by 2%.

To understand why this happens, we need to consider the relationship between current, power, and resistance.

The power (P) in an electric circuit is given by the equation P = I^2 * R, where I is the current and R is the resistance.

If the current decreases by 1%, then the new current (I`) is approximately 99% of the original current (I). In other words, I` = 0.99I.

Now, let`s calculate the new power (P`) using the new current in the equation P = I^2 * R:

P` = (0.99I)^2 * R = 0.9801 * I^2 * R

Comparing P` to the original power P, we can see that P` is approximately 98.01% of P.

Therefore, the power decreases by approximately 2%, which is option 3.

Please note that this calculation assumes that the resistance (R) remains constant. If the resistance were to change as well, then the power decrease might be different.

Hence, the correct answer is option 3