For which of these numbers is there a natural number n that makes the number end in the digit 0?

Step 1. The primes of each base. $8 = 2^3$, $15 = 3 \times 5$, $20 = 2^2 \times 5$, $14 = 2 \times 7$.

Step 2. Raise to the power $n$. The primes stay the same: $8^n$ has only $2$; $15^n$ has $3$ and $5$; $20^n$ has $2$ and $5$; $14^n$ has $2$ and $7$.

Step 3. Decide. Only $20^n$ has both a $2$ and a $5$. By the uniqueness of prime factorisation, the others can never gain the missing prime.