= Solution
For the intended standard <Hopf link>, identify $S^3$ with the unit sphere in $\mathbb C^2$ and its two components with the coordinate circles $z_1=0$ and $z_2=0$. Outside small neighborhoods of those circles, both coordinates are nonzero, and their arguments together with $|z_1|$ give
$$
S^3\setminus\operatorname{int}N_L\cong T^2\times I.
$$
On this product the longitude of either component is the meridian of the other. Therefore the two attaching meridians become the two coordinate curves of $T^2$, intersecting exactly once. Adding the product collar does not change a gluing of <solid tori>. The meridian-intersection-one argument in Question 2(c) identifies the result with the standard genus-one splitting of the sphere. Thus <zero surgery on the Hopf link> gives
$$
\boxed{M\cong S^3.}
$$
There is a genuine link-type qualification in the wording: merely specifying two linked but individually unknotted components does not determine the <Hopf link>. For example the two components of the torus link $T(2,4)$ are unknotted but have <linking number> $2$. Filling their preferred longitudes is zero surgery with linking matrix
$$
\begin{pmatrix}0&2\\2&0\end{pmatrix},
$$
so the resulting <first homology> is its cokernel, $\mathbb Z/2\oplus\mathbb Z/2$, rather than the zero group of $S^3$. The cokernel formula follows because zero longitudes impose twice the other meridian as the homology relations. Thus the sphere answer applies to the standard once-linked Hopf interpretation; the literal unspecified link description alone is insufficient to force a single manifold.
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