By what percentage will the illumination of the lamp decrease if the current drops by 20%?

We need to find the percentage decrease in illumination when current drops by 20%.

Since for a resistive lamp the power (and hence illumination) is proportional to $$I^2$$, we have $$P = I^2 R$$.

If the current drops by 20%, the new current becomes $$I' = 0.8I$$ and substituting into the expression for power gives $$P' = (0.8I)^2 R = 0.64 I^2 R = 0.64P$$.

Therefore, the percentage decrease is given by $$\frac{P - P'}{P} \times 100 = (1 - 0.64) \times 100 = 36\%$$.

The correct answer is Option 3: 36%.