To find the common ratio of the geometric progression with terms \( n-2 \), \( n \), and \( n+3 \), we use the property that the square of the middle term is equal to the product of the other two terms:

\(n^2 = (n - 2)(n + 3)\)

Expanding the right side gives:

\(n^2 = n^2 + 3n - 2n - 6\)

Simplifying leads to:

\(0 = n - 6 \implies n = 6\)

The terms of the progression are:

- First term: \( n - 2 = 4 \)
- Second term: \( n = 6 \)
- Third term: \( n + 3 = 9 \)

The common ratio \( r \) is calculated as follows:

\(r = \frac{n}{n-2} = \frac{6}{4} = \frac{3}{2}\)

We can verify it with the third term:

\(r = \frac{n+3}{n} = \frac{9}{6} = \frac{3}{2}\)

Thus, the common ratio of the geometric progression is: \(\boxed{\frac{3}{2}}\)