I am not very sure about the methodology. Please let me know if the logic is faulty anywhere.
We have $3$ slots.$$---$$
The middle one has to be a vowel. There are only two ways in which it can be filled: O, A.
Let us put O in the middle.$$-\rm O-$$
Consider the remaining letters: O, A, $\bf P_1$, $\bf P_2$, R, S, L. Since we will be getting repeated words when we use $\bf P_1$ or $\bf P_2$ once in the word, we consider both these letters as a single letter P. So, the letters now are O, A, P, R, S, L. We have to select any two of them (this can be done in $^6C_2$ ways) and arrange them (this can be done in $2!$ ways). So, total such words formed are $(^6C_2)(2!)=30$. Now there will be one word (P O P) where we will need both the P's. So we add that word. Total words=$31$.
Now, we put A in between.$$-\rm A-$$
Again proceeding as above, we have the letters O, P, R, S, L. Total words formed from them are $(^5C_2)(2!)=20$. We add two words (O A O) and (P A P). Thus, total words formed here=$22$
Hence, in all, total words that can be formed are $22+31=53$.
(P.S: Please edit if anything wrong.)
The only thing wrong is that the given answer includes arrangements with all three vowels consecutive. In other words, either you misread the problem, or else the person who came up with the given answer misread it (or forgot that vowel triplets were to be excluded). Your solution to the problem as you read it is correct.
Just for completeness, here's a somewhat simpler way to arrive at the answer for the problem as you read it: There are $7!\over2!2!$ ways to arrange the $7$ consonants. For each of these, there are $8$ places to put a vowel doublet, leaving $7$ places for the vowel singlet. Finally, there are $3!$ ways to pick a doublet, for a total of
$${7!\over2!2!}\cdot8\cdot7\cdot3!={56\cdot3!\cdot7!\over2!2!}$$
arrangements.
To get the "given" answer, we simply count the total number of ways to arrange the $10$ letters and then subtract the number of ways in which the $3$ vowels are all isolated:
$${10!\over2!2!}-{7!\over2!2!}\cdot8\cdot7\cdot6=(10\cdot9-7\cdot6)\cdot8\cdot{7!\over2!2!}=48\cdot8\cdot{7!\over2!2!}={64\cdot6\cdot7!\over2!2!}$$
Best Answer
First, arrange the six consonants. There are $6!$ ways to do this. There are $7$ possible places to place the $4$ vowels, with no more than one in each place, so you choose $4$ of the $7$ places.
The total number of arrangements with no consecutive vowels is the number of ways to arrange the consonants, multiplied by the number of ways to permute the vowels, multiplied by the number of ways to arrange the vowels or $6!\cdot \dfrac{4!}{2!} \times \large\binom{7}{4}$.