Answer
$f_{min,1}$ must be multiplied by $~~5$
Work Step by Step
We can find the path length difference:
$\Delta L = 19.5~m-18.3~m = 1.2~m$
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:
$f_{min,1} = \frac{v}{\lambda}$
$f_{min,1} = \frac{v}{\Delta L/0.5}$
$f_{min,1} = \frac{0.5~v}{\Delta L}$
$f_{min,1} = \frac{(0.5)~(343.~m/s)}{1.2~m}$
$f_{min,1} = 143~Hz$
We can find an expression for $f_{min,3}$:
$f_{min,3} = \frac{v}{\lambda}$
$f_{min,3} = \frac{v}{\Delta L/2.5}$
$f_{min,3} = \frac{2.5~v}{\Delta L}$
$f_{min,3} = 5\times ~\frac{0.5~v}{\Delta L}$
$f_{min,3} = 5\times ~f_{min,1}$
$f_{min,1}$ must be multiplied by $~~5$