Yeah I'd fail that.
Every time you add a new character, the existing string has to be re-created. That's the same kind of expensive as moving all the following characters each time one is removed, see the resemblance?
You'd get around it by appending to a list instead.
Other methods are using regex to substitute away what matches a vowel pattern, or using a generator expression which filters out vowels.
The list method you certainly can implement, the other two go like:
return re.sub('[aoeuiAOEUI]', '', text)
return ''.join(chr for chr in text if chr not in 'aoeuiAOEUI')
There are plenty of other ways, for example splitting the text wherever there is a vowel, and then joining it:
return ''.join(re.split('[aoeuiAOEUI]', text))
re.sub is probably the best one overall, the generator expression is also very nice though, it reads nearly like English. Splitting and joining is a little silly, doing more than required, but sure. It'll do the job.