Answer
From speaker #2, this point is a distance of $2.61~m$, $3.27~m$, or $3.93~m$.
Work Step by Step
We can find the wavelength:
$\lambda~f = v$
$\lambda = \frac{v}{f}$
$\lambda = \frac{343~m/s}{523~Hz}$
$\lambda = 0.66~m$
Note that $\frac{\lambda}{2} = \frac{0.66~m}{2} = 0.33~m$
In order for destructive interference to occur, the path difference between the two speakers must have the form $(n~\lambda +\frac{\lambda}{2})$ for some integer $n$. Therefore, the possible distances from speaker #2 must have the form $2.28~m + (n~\lambda+\frac{\lambda}{2})$. We can find the possible distances from speaker #2:
$d_0 = 2.28~m+(0)(0.66~m)+0.33~m = 2.61~m$
$d_1 = 2.28~m+(1)(0.66~m)+0.33~m = 3.27~m$
$d_2 = 2.28~m+(2)(0.66~m)+0.33~m = 3.93~m$
From speaker #2, this point is a distance of $2.61~m$, $3.27~m$, or $3.93~m$.
