Suppose the foliation has codimension k. Should the transversal that you are asking for have codimension k as well? If yes then there are multiple possibilities for the topology of a transversal.

Thanks for the nice answer. Just to make sure, your answer gives infinitely many distinct algebraic integers with the same discriminant, right? Would you mind mentioning the name of the reference in the sentence `See here, page 10 of Section 4, for a good reference'? It seems the page link does not work anymore.

Just to clarify, as Ryan mentioned, the above argument only shows that the diffeomorphism type of the resulting manifold is uniquely determined by the image of the meridian up to isotopy.

If we want to consider the case of compact, orientable 3-manifolds then the statement should be modified as follows: every compact orientable 3-manifold can be obtained by removing a regular neighborhood of some graph G in S^3 and integral surgery on a link L in S^3 - G. In any case, for the above argument we only need to consider a specific boundary component of M_1 which coincides with the boundary of B^3, and the argument is as before.

Assume for the moment that we are talking about closed orientable 3-manifolds. Then the boundaries of B^3 and M_1 coincide since after gluing them together along their boundary, we obtain a closed manifold.

@IgorBelegradek Thanks for pointing it out. I just edited the question.

@IgorBelegradek I think the second reference is studying principal circle bundles [so the group of the bundle is S^1] whereas my question is about smooth circle bundles [so the group of the bundle is Diff(S^1)]. The if and only if statement is not true for circle bundles which are not necessarily principal circle bundles; I'm sure you are aware since you referred to MW inequality in your answer.

Thanks, Igor. The reference you mentioned is great (a paper of Miyoshi from 2001 whose content is the proof of the above remark). Do you have any guess for the answer to the second question?