The median of first six prime numbers is :

1. 5

2. 7

3. 6

4. 7.5

Prime numbers : 2, 3, 5, 7, 11, 13.

No. of terms (n) = 6, which is even.

Median = $\dfrac{\Big(\dfrac{n}{2}\Big) \text{ th term} + \Big(\dfrac{n}{2} + 1\Big) \text{ th term}}{2}$

Substituting values we get :

$\text{Median } = \dfrac{\Big(\dfrac{6}{2}\Big) \text{ th term} + \Big(\dfrac{6}{2} + 1\Big) \text{ th term}}{2} \\[1em] = \dfrac{\text{3rd term + 4th term}}{2} \\[1em] = \dfrac{5 + 7}{2} \\[1em] = \dfrac{12}{2} \\[1em] = 6.$

Hence, Option 3 is the correct option.