$2$ must be the last digit because $1, 2$, and $3$ are not divisible by 2, and not even. So the simple rules of how many ways we can arrange things comes into place. Because there are 3 other digits, the way we can arrange the 4 digits $2$ being the last one is $3!$.
$3! = 6$, 6 4 digit even numbers.
The reason $3!$ works is because $3! = 3\times 2 \times 1$, and we have 3 digits, $1,2,3$. All 3 numbers can always be the first number. If one number is first, there are 2 numbers left to be the second digit, and if 2 numbers are digits 1 and 2, only one number is left to be the last digit, hence $3 \times 2 \times 1$
This property applies to any chain of $n$ length. The amount of ways we cabn arrrange the chain $= n \times (n-1) \times (n-2) \ \ldots \times 2 \times 1$, or $n!$
As this applies to $VIRTUAL$, it can be expressed as a chain of length 5 because the first and last letters never change. $5! = 120$