A 23 kHz sine wave sampled at 48 kHz, then FFT
Since you used 2.3k twice, I am assuming you meant that with the original 23k value.

Converting your frequency:

$$ 2.3k \frac{cycles}{second} \cdot \frac{ 64 \frac{samples}{frame} }{ 48k \frac{samples}{second}  }  \approx 3.07 \frac{cycles}{frame}$$

Because this is not an integer value, the peak DFT value will be at bin 3 and there will be values in all the other bins, which is commonly called "leakage".  Being a real valued signal, the upper half of the DFT will be a mirror image with complex congjugate values, so there will also be a peak at bin 61 (64-3).

Your 100 sample figure is irrelevant if you are only using 64 of them for the DFT.

Signals have to be a whole number of cycles per frame in order for the DFT not to have leakage.  The other answers presume you understand that the DFT is a dirac delta, windowed, convolved version of the continuous FT.  This is not an easy way to understand it, but it is how it is usually taught.  It also leads to the misconception that the leakage is described by what is known as the sinc function $ \frac{\sin(d)}{d} $.  That is not true.  The fall off for a complex valued signal is described by what is known as the aliased sinc function $ \frac{\sin(Nd)}{N\sin(d)} $.  Another name is "Dirichlet function".  A real valued signal is actually the sum of two complex signals, so that function doesn't apply either.

You can find the exact formula for the bin values, the peak and "leakage", in my blog article "DFT Bin Value Formulas for Pure Real Tones".

You can also find a much more intuitive explanation of why leakage occurs in another one of my blog articles "DFT Graphical Interpretation: Centroids of Weighted Roots of Unity".

Hope this helps,

Ced