The correct answer is A. Decreased by 2%.
The intensity of illumination on the screen is inversely proportional to the square of the distance between the projector and the screen. This means that if the distance is increased by 2%, the intensity of illumination will be decreased by 4%.
To understand this, let’s consider the following example. Suppose the distance between the projector and the screen is 100 meters. The intensity of illumination on the screen will be $I_0$. If the distance is increased to 102 meters, the intensity of illumination will be $I_1$. We can calculate $I_1$ as follows:
$$I_1 = \frac{I_0}{(102)^2} = \frac{I_0}{10404} = 0.09607$
This is a decrease of 4% from $I_0$.
In general, if the distance between the projector and the screen is increased by $x$%, the intensity of illumination will be decreased by $2x$%.