Answer
$f_{min,1} = 343~Hz$
Work Step by Step
We can find the distance from the upper speaker to the listener:
$d = \sqrt{(2.00~m)^2+(3.75~m)^2} = 4.25~m$
We can find the path length difference:
$\Delta L = 4.25~m-3.75~m = 0.50~m$
To produce fully destructive interference, $\frac{\Delta L}{\lambda} = 0.5, 1.5, 2.5,...$
Then: $\lambda = \frac{\Delta L}{0.5}, \frac{\Delta L}{1.5},\frac{\Delta L}{2.5},...$
We can find the lowest frequency that produces fully destructive interference:
$f_{min,1} = \frac{v}{\lambda}$
$f_{min,1} = \frac{v}{\Delta L/0.5}$
$f_{min,1} = \frac{0.5~v}{\Delta L}$
$f_{min,1} = \frac{(0.5)~(343.~m/s)}{0.50~m}$
$f_{min,1} = 343~Hz$