Let f = the 1st digit
Then the last digit = 2f
So the number looks like: f t t 2f
Doubling the first two digits gives the last two. Treat each as a two digit number.
Then we can write this in equation form as
2(10f + t) = 10t + 2f
20f + 2t = 10t + 2f
18f = 8t
9f = 4t
t = (9/4)f
Now the number looks like: f (9/4)f (9/4)f 2f
For the digits to be integers from 0 to 9, f must equal 4. Any other value gives a non-integer digit or a digit > 9.
So the solution is 4 9 9 8