Oh, this is an interesting question...
At first, I took at look at it and thought the same thing as you; I also thought that the frequency should be based off the formula f = nv/4L for strings.
It turns out there are actually many different cases for strings:
Case 1. 1D - 2 Fixed ends: Lambda = 2L/n and f = nv/2L
Case 2. 1D - 2 Free ends: Lambda = 2L/n and f = nv/2L
Case 3: 1D - 1 Fixed end and 1 Free end: Lambda = 4L/n and f = nv/4L
In this problem, it would be Case 3 since one end is attached to the peg. You can use the formula Lambda = 4L/n; since we know that the string only goes through 3/4 of the wavelength, you can multiply both sides by 3/4 to give (3/4)Lambda = 3L/n and you know the length L is fixed, (3/4)Lambda must still equal L (even though it only goes through 3/4 of the wavelength, it still uses the whole length L) so n = 3. Another way to think about it is to just rearrange the formula to give n=4L/Lambda which will give you the sequence that
@Cawolf mentioned
Hope that makes sense! This was actually a great problem to think through and helped me to understand the situation better too
🙂 Thanks! Where did you find these questions?