In a Vernier caliper, n divisions of the Vernier scale coincide...

Let the number of Vernier scale divisions be $n$.

Given that $n$ Vernier scale divisions (VSD) coincide with $(n-3)$ main scale divisions (MSD).

$n \text{ VSD} = (n-3) \text{ MSD}$
$1 \text{ VSD} = \dfrac{n-3}{n} \text{ MSD}$

The least count (L.C.) is given by:
$\text{L.C.} = 1 \text{ MSD} - 1 \text{ VSD}$
$\text{L.C.} = 1 \text{ MSD} - \dfrac{n-3}{n} \text{ MSD} = \dfrac{3}{n} \text{ MSD}$

Substitute the given values ($\text{L.C.} = 0.06 \text{ mm}$ and $1 \text{ MSD} = 0.5 \text{ mm}$):
$0.06 = \dfrac{3}{n} \times 0.5$
$0.06 = \dfrac{1.5}{n}$
$n = \dfrac{1.5}{0.06} = 25$

Answer: $25$