Solution:

Let $A$ and $B$ be the endpoints of the tunnel, and let $C$ be the point diametrically opposite to $A$. We claim that the plane $\pi$ which perpendicularly bisects $BC$ gives the desired cut. Indeed, this plane contains the center of the apple since it is equidistant from $B$ and $C$; now assume for the sake of contradiction that the tunnel enters both halves of the apple. Then it must cross $\pi$ at some point $R$. Reflect that portion of the tunnel lying between $R$ and $B$ across $\pi$, thus obtaining a tunnel from $R$ to $C$. Note that reflection preserves the length of this segment of tunnel. Thus we obtain a tunnel from $A$ to $R$ to $C$, which has the same length as the original $(61~\mathrm{mm})$. On the other hand, since $A$ and $C$ are diametrically opposite, by the triangle inequality the tunnel has length $\geq AC = 62~\mathrm{mm}$. This is a contradiction. Thus our tunnel cannot intersect both halves, which is what we want.