The pulling force must counteract the variable weight of the bucket plus the hanging portion of the rope. Let $x$ be the height of the bucket from the bottom of the well. The length of the rope still hanging is $(H - x)$.

The required pulling force $F(x)$ at height $x$ is:

$$F(x) = (M + m_{rope})g = (M + \lambda(H-x))g$$

Since the force is a linear function of displacement $x$, the work done $W$ can be calculated as the area of a trapezoid under the force-displacement graph from $x=0$ to $x=H$.

The initial force at the bottom ($x=0$) is $F_i = (M + \lambda H)g$. The final force at the top ($x=H$) is $F_f = Mg$.

The work done is the average force multiplied by the total displacement $H$:

$$W = \bar{F} H = \frac{F_i + F_f}{2} H$$ $$W = \frac{(M + \lambda H)g + Mg}{2} H = \left(M + \frac{1}{2}\lambda H\right)gH$$

Substituting the given values, with $\lambda = 0.2$ kg/m and $g = 9.8$ m/s$^2$:

$$W = \left(20 \text{ kg} + \frac{1}{2}(0.2 \text{ kg/m})(10 \text{ m})\right)(9.8 \text{ m/s}^2)(10 \text{ m})$$ $$W = (20 + 1)(98) \text{ J} = 21 \times 98 \text{ J}$$