For what frequencies does their sound at the speakers produce constructive interference?

You need to find the path difference. That is, how much further must sound waves from the more distant speaker travel than the close speaker, to reach the mike.

Use Pythagoras to find the distance of the further speaker: it is √ (2.00²+4.50²) = 4.924m so the path difference is 4.924-4.50=0.424m.

You will get constructive interference when this path difference is an integer number of wavelengths, because the waves will arrive at the mike in phase.

The speed of sound is 340m/s so the lowest frequency that will produce an antinode at the mike is the one that makes 0.424=λ

v=fλ so f=v/λ

f=340/0.424=801Hz.

The next one will be when 0.424m = 2λ = > λ=0.212m

f=340/0.212=1602Hz

and so-on according to f=340n/0.424 where n is an integer.

For destructive interference the path difference must be (n-½) λ because that will make the waves arrive at the mike 180° out of phase.

f=340 (n-½) / 0.424