For an another way using probability:
What is the probability that $a,a,u,u,e,o$ will be in the form of correct alphabetical order ? The answer is : $$\frac{1}{6!/(2!2!)}=\frac{1}{180}$$
Now, when we order the letters of "Polyunsaturated", $1/180$ of these arrangements will be in the form that vowels are in alphabetical order.
Then , $$\frac{15!}{2!2!2!}\times \frac{1}{180}=908,107,200$$
For the third approach:
The word "Polyunsaturated" has $15$ letters, so we can think that there are $15$ empty places for replacing these letters.Firstly, lets select $6$ place for vowels among these $15$ places by $C(15,6)$ ways. When we select these $6$ places, the letters can be replaced in only one way such that they are in the form of alphabetical order.After that, we now have $9$ places to place these $9$ constonants.We can order them $9!/2!$ ways in a line.Then, $$\binom{15}{6}\times \frac{9!}{2!}=5005 \times 181,440=908,107,200$$