Answer
$41.22\;m$
Work Step by Step
In a single-slit diffraction pattern, if the dark fringes (minima) above and below the central axis can be located with the general equation
$a\sin\theta=m\lambda,\;\;\;\;\;\text{for}\; m=1,2,3,...$
where $a$ is slit width, $\lambda$ is the wavelength and $\theta$ is the angle of the minima with respect to the central axis.
In this problem, sound waves with frequency $3000\,Hz$ and speed $343\;m/s$ diffract through the rectangular opening of a speaker cabinet and into a large auditorium of length $d=100\;m$. The opening has a horizontal width of $a=30.0\;cm=0.3\;m$
Therefore, the wavelength of the sound wave is given by
$\lambda=\frac{343}{3000}\;m$
Let the first minimum is at a distance $y$ from the central axis. Then
$\sinθ=\frac{y}{\sqrt {d^2+y^2}}$
For first minimum, we can write
$a\sin\theta=\lambda$
or, $\sin\theta=\frac{\lambda}{a}$
or, $\frac{y}{\sqrt {d^2+y^2}}=\frac{\lambda}{a}$
$y^2\times \frac{a^2}{\lambda^2}=d^2+y^2$
or, $y= \frac{d}{\sqrt{\frac{a^2}{\lambda^2}-1}}$
or, $y= \frac{100}{\sqrt{\big(\frac{0.3\times 3000}{343}\big)^2-1}}$
or, $y=41.22\;m$
Therefore, a listener at a distance $41.22\;m$ from the central axis will be at the first diffraction minimum.
