A uniform cable of mass $$M$$ and length $$L$$ is placed on a horizontal surface such that its $$\left(\frac{1}{n}\right)^{th}$$ part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be:

Let us begin by noting that the cable is uniform, so its mass per unit length (linear mass density) is constant. We denote this density by $$\lambda$$ and write

$$\lambda \;=\;\frac{M}{L},$$

because the total mass $$M$$ is uniformly distributed along the total length $$L$$.

According to the statement, a fraction $$\left(\frac{1}{n}\right)^{\!th}$$ of the cable hangs below the horizontal surface. Hence the vertical length that is hanging is

$$\ell_h \;=\;\frac{L}{n}.$$

We choose the horizontal surface as the reference level $$y = 0$$. Positive $$y$$ is taken upward, so the hanging portion originally lies at negative values of $$y$$. Every small element of cable located at a distance $$y$$ below the surface (that is, at $$y$$ between $$-\,\frac{L}{n}$$ and $$0$$) has to be raised through a height equal to the absolute value of its initial coordinate, namely $$|y| = -y$$.

Let us label a small element of length $$dy$$ on the hanging part. Its mass is

$$dm \;=\;\lambda\,dy \;=\;\frac{M}{L}\,dy.$$

Initially this element is at a coordinate $$y$$ (negative), so it must be lifted through a vertical distance $$h = -y$$ to reach the surface. The elementary work $$dW$$ required to lift this element is given by the basic work formula

$$dW \;=\;(\text{force})\times(\text{displacement}) \;=\; (dm\,g)\,h.$$

Substituting $$dm$$ and $$h$$, we have

$$dW \;=\;\left(\frac{M}{L}\,dy\right)g\,(-y) \;=\;-\frac{M g}{L}\,y\,dy.$$

Because $$y$$ runs from $$-\frac{L}{n}$$ (bottom of the hanging part) up to $$0$$ (surface), the total work $$W$$ is obtained by integrating $$dW$$ over this interval:

$$W \;=\;\int_{y=-L/n}^{\,0} \left(-\frac{M g}{L}\,y\right)\,dy.$$

We now carry out the algebraic steps of the integration:

$$W \;=\;-\frac{M g}{L}\int_{-L/n}^{\,0} y\,dy \;=\;-\frac{M g}{L}\left[\,\frac{y^{2}}{2}\,\right]_{-L/n}^{\,0}.$$

Evaluating the definite integral, we get

$$\left[\,\frac{y^{2}}{2}\,\right]_{-L/n}^{\,0} \;=\;\frac{(0)^{2}}{2} \;-\;\frac{\left(-\dfrac{L}{n}\right)^{2}}{2} \;=\;0 \;-\;\frac{L^{2}}{2n^{2}} \;=\;-\frac{L^{2}}{2n^{2}}.$$

Substituting this back,

$$W \;=\;-\frac{M g}{L}\left(-\frac{L^{2}}{2n^{2}}\right) \;=\;\frac{M g}{L}\,\frac{L^{2}}{2n^{2}} \;=\;\frac{M g L}{2n^{2}}.$$

Thus the work needed to raise the entire hanging part of the cable to the level of the surface is

$$W \;=\;\frac{M g L}{2 n^{2}}.$$

Comparing this result with the given options, we see that it matches Option A.

Hence, the correct answer is Option A.