Incomplete question as we have not told to find which quantity.The complete question is here
A flutist assembles her flute in a room where the speed of sound is 340 m/s. When she plays the note A, it is in perfect tune with a 440 Hz tuning fork. After a few minutes, the air inside her flute has warmed to where the speed of sound is 347 m/s .
(a)How many beats per second will she hear if she now plays the note A as the tuning fork is sounded?
(b) How far does she need to extend the "tuning joint" of her flute to be in tune with the tuning fork?
Answer:
(a)
(b) The flutist extends to the turning joint of her flute by 1.363×10⁻⁵m
Explanation:
Part A
The wavelength of sound wave of the note A
λ=v/f
Where v speed of sound
f is frequency
Substitute the given values
So
λ=(340m/s)÷440Hz
λ=0.7727 m
The new frequency of note A when the air inside the flute has warmed
f¹=v¹/λ
f¹=(347m/s)/0.7727 m
f¹=449.05Hz
The Beat frequency of two waves is:

Part B
Frequencies of standing waves modes of an open tube of Length L:

So

L=mλ/2
From part (a) The wavelength of sound of note A is 0.7727m and m=1
So

When the air inside the flute has warmed.Then the new length is given below at same frequency of 440 Hz
So

So the length the flutist extends to the turning joint for the flute to be:
ΔL=La-L
ΔL=0.38636m - 0.38635
ΔL=1.363×10⁻⁵m
The flutist extends to the turning joint of her flute by 1.363×10⁻⁵m