Hydrolysis of epoxide

[![enter image description here](https://i.stack.imgur.com/ysqcY.jpg)](https://i.stack.imgur.com/ysqcY.jpg)


There are two possibilities:


* Oxonium ion formation
* Cleaving of epoxide


Firstly, I don't know if epoxide is more reactive than carbonyl group.
If epoxide reacts with the proton,
[![enter image description here](https://i.stack.imgur.com/MOIOC.jpg)](https://i.stack.imgur.com/MOIOC.jpg)


But the given answer is:


[![enter image description here](https://i.stack.imgur.com/LnlYl.png)](https://i.stack.imgur.com/LnlYl.png)


In the above mechanism, an unstable carbocation is formed because its near a carbonyl group which is electron withdrawing:


[![enter image description here](https://i.stack.imgur.com/raq0n.jpg)](https://i.stack.imgur.com/raq0n.jpg)




I would go with your approach but your final product is unstable due to high repulsion's by $\pi$ electrons. 


[![enter image description here](https://i.stack.imgur.com/JuobPm.png)](https://i.stack.imgur.com/JuobPm.png)


Hence your final product will react further to form a more thermodynamically feasible product.


[![enter image description here](https://i.stack.imgur.com/jIBTT.png)](https://i.stack.imgur.com/jIBTT.png)


NOTES:  

In second structure two anti-periplanar migrations are possible and the one which leads to the more stable carbonation is favored (or consider the better migrating group based on electronegativity). Third structure rotates about spiral carbon due to repulsion's. 


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