To solve this problem, we need to determine the work done to lift the hanging part of the cable up to the surface. The work done lifting a small element of the cable will be the weight of the element times the distance it has to be lifted.

Let's take a small section of the cable at a depth $x$ below the surface. This section has a length of $dx$, so its mass is $(M/L)dx$ where $(M/L)$ is the linear mass density of the cable.

The work $dW$ done to lift this small section up to the surface is the weight of the section times the distance it has to be lifted :

$dW = (M/L)gdx \times x$.

Integrating this expression from 0 to L/n (the length of the hanging part of the cable) gives the total work done :

$$W = \int\limits_{0}^{L/n} (M/L)gxdx$$

$$= (Mg/L) \int\limits_{0}^{L/n} xdx$$

$$= (Mg/L) \times [x^2/2]_{0}^{L/n}$$

$$= (Mg/L) \times [L^2/(2n^2)]$$

$$= MgL/(2n^2)$$

So the work done to lift the hanging part of the cable up to the surface is $MgL/(2n^2)$.

Therefore, the correct answer is Option C :

$$\frac{MgL}{2n^2}$$