At first, looking at 2TQFT I see no reason to expect it to be locally small. But we know that 2TQFT is equivalent to the category of commutative Frobenius algebras cFA$_{\mathbb K}$, which is locally small. But is this enough to conclude that 2TQFT is also locally small? Certainly the skeleton of 2TQFT is, being isomorphic to cFA$_{\mathbb K}$...
I am new to these grounds, so I am a bit unsure of my steps.