Answer
$7.6\;\rm m$
Work Step by Step
The crack in the cave is representing a single-slit situation where the crack width is the single-slit.
We need to find the width of the sound beam at 100 m from the crack which is similar to the width of the central maximum fringe in a single-slit experiment.
We know that the width of this central maximum is given by
$$w=\dfrac{2\lambda L}{a}$$
where $\lambda=v/f$
$$w=\dfrac{2v L}{af}$$
Plugging the known;
$$w=\dfrac{2(340)(100)}{(0.30)(30\times 10^3)}$$
$$w=\color{red}{\bf 7.6}\;\rm m$$
