First separating the word estate into vowels and non vowels gives
$EAE$ for the vowels and $S,T,T$ for the non-vowels. I interpreted this as a group of 4 where there's two T's resulting in $\frac{4!}{2!}$ but since the vowels can also be permuted and there's two E's this results in
#permutations = $\frac{4!}{2!}\cdot\frac{3!}{2!}=3!3!=36$.
The correct answer is $180$ so I'm off by a factor of 5 although I have no clue where it comes from. If someone could point out the flaw in my reasoning that would be great.
Best Answer
Your result is correct. If all the vowels are adjacent we have three possible vowel-blocks: $AEE$, $EAE$ and $EEA$. Then the total number of arrangents of $S$, $T$, $T$ and the vowel-block ($4$ elements with a double letter) is $$3\cdot \frac{4!}{2!}=36$$
Note that the total number of anagrams of the word $ESTATE$ with no restrictions is $$\frac{6!}{2!2!}=180.$$