B

Given that the spherical ball has a radius $r$ and relative density 0.5, we can start calculating the work done in pushing the ball down so that it is fully immersed in water.

**Step 1: Find the volume of the sphere**
The volume of the sphere, $V$, is given by:\
$ V = \frac{4}{3} \pi r^3 $

**Step 2: Calculate the weight of the sphere**
The weight of the sphere, $W_s$, is given by its volume times the density (density of water is $\rho$):\
$ W_s = V \times (0.5 \cdot \rho) = \frac{4}{3} \pi r^3 \cdot (0.5 \cdot \rho) = \frac{2}{3} \pi r^3 \rho $

**Step 3: Calculate the buoyant force when fully immersed**
When the sphere is fully immersed, the buoyant force, $F_b$, is equal to the weight of the water displaced by the volume of the sphere. This is given by:\
$ F_b = V \times \rho = \frac{4}{3} \pi r^3 \cdot \rho $

**Step 4: Calculate the net force when the ball is fully immersed**
When the ball is fully immersed, the net upward force, $F_{net}$, is determined by the difference between the buoyant force and the weight of the sphere:

$ F_{net} = F_b - W_s = \frac{4}{3} \pi r^3 \cdot \rho - \frac{2}{3} \pi r^3 \cdot \rho = \frac{2}{3} \pi r^3 \cdot \rho $

**Step 5: Work done in pushing the sphere down**
The work done, $W$, is equal to the force multiplied by the distance over which the force is applied. In this case, the distance is the radius $r$ (since we are immersing the sphere fully from its equilibrium position):

$ W = F_{net} \cdot \text{distance} = \frac{2}{3} \pi r^3 \cdot \rho \cdot r = \frac{2}{3} \pi r^4 \cdot \rho $

Identifying which option corresponds to this solution gives us option B. Thus, the work done in pushing the ball down so that it is fully immersed is option B.