Answer: $C$
The given expression can be simplified as:
$$=\bigg| e^{\frac{2\pi ik}{n}}- e^{\frac{2\pi i(k-1)}{n}}\bigg| $$
$$= \bigg| e^{\frac{2\pi i(k-1)}{n}}\bigg |\bigg |e^{\frac{2\pi i}{n}}-1\bigg | $$
$$= \bigg|\cos\bigg (\frac{2\pi}{n}\bigg )-1 + i\sin\bigg (\frac{2\pi}{n}\bigg )\bigg |$$
$$= \bigg | -2\sin^2\bigg (\frac{\pi}{n}\bigg) + 2i\sin\bigg(\frac{\pi}{n}\bigg)\cos\bigg (\frac{\pi}{n}\bigg)\bigg |$$
$$= 2\sin\bigg(\frac{\pi}{n}\bigg )$$
Now, the required limit will be:
$$\therefore \underset{n \to \infty }{\lim} \sum^{n}_{k = 1}2\sin \bigg (\frac{\pi}{n}\bigg ) = \underset {n \to \infty}{\lim}2n\sin\bigg (\frac{\pi}{n}\bigg ) = 2\pi$$
$\therefore$ C is the right answer.