1. Given data

• Number of small drops: $n = 64$
• Radius of each small drop: $r = 0.02\,\text{m}$
• Charge on each small drop: $q = 5\,\mu\text{C}$

We need the ratio of surface charge density of the bigger drop to that of a smaller drop.

2. Surface charge density of a conducting sphere

For a conducting spherical drop,

$\sigma = \frac{Q}{4\pi R^2}$

where $Q$ is charge and $R$ is radius.

3. Radius of the bigger drop

When $64$ identical drops combine, volume is conserved:

$\frac{4}{3}\pi R^3 = 64 \cdot \frac{4}{3}\pi r^3$

So,

$R^3 = 64r^3$

$R = 4r$

4. Charge on the bigger drop

Charges add up:

$Q = 64q$

5. Surface charge density of small drop

$\sigma_s = \frac{q}{4\pi r^2}$

6. Surface charge density of bigger drop

$\sigma_b = \frac{Q}{4\pi R^2} = \frac{64q}{4\pi (4r)^2}$

$\sigma_b = \frac{64q}{4\pi \cdot 16r^2} = \frac{4q}{4\pi r^2}$

7. Required ratio

$\frac{\sigma_b}{\sigma_s} = \frac{\frac{4q}{4\pi r^2}}{\frac{q}{4\pi r^2}} = 4$

Hence,

$\sigma_b : \sigma_s = 4 : 1$

8. Option check

• A: $1:4$ ❌
• B: $4:1$ ✅
• C: $1:8$ ❌
• D: $8:1$ ❌

So the correct answer is B.