1. What is being asked?

   We need the ratio $\frac{\text{electronic charge measured on the moon}}{\text{electronic charge measured on the earth}}$ using Millikan’s oil drop experiment.

2. Key idea of Millikan’s oil drop experiment

   In Millikan’s experiment, the charge on a drop is determined by balancing electric force with gravitational effects. Although the calculation involves $g$, the final measured fundamental charge $e$ is a universal constant.

   The electronic charge does not depend on location (earth or moon). Only the experimental conditions, such as weight of the oil drop, change because gravity changes.

3. Effect of changing gravity

   On the moon, acceleration due to gravity is $g_M$; on earth, it is $g_E$.

   This changes the weight of the oil drop: $W = mg$ so the balancing electric field needed would be different.

   But the value of the fundamental charge obtained from the experiment remains the same: $e_{\text{moon}} = e_{\text{earth}}$

4. Required ratio

   Therefore, $\frac{e_{\text{moon}}}{e_{\text{earth}}} = 1$

5. Option check

   • A: $\dfrac{g_M}{g_E}$ — incorrect
   • B: $1$ — correct
   • C: $0$ — incorrect
   • D: $\dfrac{g_E}{g_M}$ — incorrect

Therefore, the correct answer is B.