Explanation

This looks like an arithmetic sequence problem.
If the first term $a = 3$ and last term $l = 18$, and there are 4 numbers in between, the total terms $n=6$ (3 + 4 + 18).

The terms:
$a_1 = 3$
$a_6 = 18$
Number of terms $n = 6$

Common difference $d = frac{18 - 3}{6 - 1} = frac{15}{5} = 3$

The 3rd middle number is the 4th term:
$a_4=a_1+3d=3+3\times 3=12$