The correct answer is $\boxed{\text{C. }\frac{1}{2}\sum\limits_{r = 0}^{N – 1} {X\left( r \right)X\left( {k + r} \right)}}$.
The DFT of a sequence $x(n)$ is defined as
$$X(k) = \sum_{n=0}^{N-1} x(n) e^{-j2\pi nk/N}$$
where $N$ is the length of the sequence.
The sequence $y(n)$ is given by
$$y(n) = \frac{1}{N}\sum_{r=0}^{N-1} x(r) x(n+r)$$
The DFT of $y(n)$ is given by
$$Y(k) = \frac{1}{N}\sum_{n=0}^{N-1} y(n) e^{-j2\pi nk/N}$$
Substituting the expression for $y(n)$ into the above equation, we get
$$Y(k) = \frac{1}{N}\sum_{n=0}^{N-1} \frac{1}{N}\sum_{r=0}^{N-1} x(r) x(n+r) e^{-j2\pi nk/N}$$
$$= \frac{1}{N^2}\sum_{r=0}^{N-1} \sum_{n=0}^{N-1} x(r) x(n+r) e^{-j2\pi nk/N}$$
$$= \frac{1}{N^2}\sum_{r=0}^{N-1} x(r) \left(\sum_{n=0}^{N-1} x(n) e^{-j2\pi rn/N}\right) e^{-j2\pi nk/N}$$
$$= \frac{1}{N^2}\sum_{r=0}^{N-1} x(r) X(r) e^{-j2\pi nk/N}$$
$$= \frac{1}{N}\sum_{r=0}^{N-1} X(r) e^{-j2\pi nk/N} \left(\frac{x(r)}{N}\right)$$
$$= \frac{1}{N}\sum_{r=0}^{N-1} X(r) e^{-j2\pi nk/N} \left(\frac{1}{N}\sum_{n=0}^{N-1} x(n) e^{j2\pi nr/N}\right)$$
$$= \frac{1}{N}\sum_{r=0}^{N-1} X(r) e^{-j2\pi nk/N} \left(X(r)\right)^*$$
$$= \frac{1}{2}\sum_{r=0}^{N-1} X(r) X(k+r)$$
Therefore, the DFT of the sequence $y(n)$ is given by
$$Y(k) = \frac{1}{2}\sum_{r=0}^{N-1} X(r) X(k+r)$$