To solve this problem, we can use the concepts of terminal velocity and the forces acting on the spherical ball. First, let's analyze the situation step-by-step.

When the ball falls freely under gravity, it achieves a terminal velocity $$v_t$$ in water. This terminal velocity is reached when the gravitational force is balanced by the drag force and the buoyant force in the water.

The forces acting on the ball are:

1. Gravitational Force: $$F_g = mg$$

2. Buoyant Force: $$F_b = \rho_{\text{water}} V g$$

3. Drag Force: $$F_d = 6 \pi \eta r v_t$$

Where,

$$m$$ is the mass of the ball.

$$g$$ is the acceleration due to gravity ($$9.8 \, \mathrm{m/s^2}$$).

$$\rho_{\text{water}}$$ is the density of water ($$1000 \, \mathrm{kg/m^3}$$).

$$V$$ is the volume of the ball ($$\frac{4}{3} \pi r^3$$).

$$\eta$$ is the coefficient of viscosity of water ($$9.8 \times 10^{-6} \, \mathrm{Ns/m^2}$$).

$$r$$ is the radius of the ball ($$1 \times 10^{-4} \, \mathrm{m}$$).

$$v_t$$ is the terminal velocity.

Using the equilibrium condition at terminal velocity:

$$F_g = F_b + F_d$$

$$mg = \rho_{\text{water}} V g + 6 \pi \eta r v_t$$

First, compute the mass of the ball:

$$m = \rho_{\text{ball}} \times V = \rho_{\text{ball}} \times \frac{4}{3} \pi r^3$$

$$m = 10^5 \, \mathrm{kg/m^3} \times \frac{4}{3} \pi (1 \times 10^{-4} \, \mathrm{m})^3$$

$$m = 10^5 \,\mathrm{kg/m^3} \times \frac{4}{3} \pi \times 10^{-12} \,\mathrm{m^3}$$

$$m = \frac{4}{3} \pi \times 10^{-7} \, \mathrm{kg}$$

Next, solve for the terminal velocity $$v_t$$ using the equilibrium equation:

$$mg = \rho_{\text{water}} \frac{4}{3} \pi r^3 g + 6 \pi \eta r v_t$$

$$v_t = \frac{mg - \rho_{\text{water}} \frac{4}{3} \pi r^3 g}{6 \pi \eta r}$$

$$v_t = \frac{\frac{4}{3} \pi \times 10^{-7} \times 9.8 - 1000 \times \frac{4}{3} \pi (1 \times 10^{-4})^3 \times 9.8}{6 \pi \times 9.8 \times 10^{-6} \times 10^{-4}}$$

$$v_t = \frac{\frac{4}{3} \pi \times 10^{-7} \times 9.8 - 1000 \times \frac{4}{3} \pi \times 10^{-12} \times 9.8}{6 \pi \times 9.8 \times 10^{-10}}$$

$$v_t = \frac{\frac{4}{3} \pi \times 9.8 \times 10^{-7} (1 - 10^{-5})}{6 \pi \times 9.8 \times 10^{-10}}$$

$$v_t = \frac{\frac{4}{3} \times 10^{-7}}{6 \times 10^{-10}}$$

$$v_t = \frac{4}{18} \times 10^3 \, \mathrm{m/s}$$

$$v_t \approx 222.22 \, \mathrm{m/s}$$

The height $$h$$ required to reach this terminal velocity while the ball falls freely under gravity can be found using the kinematic equation:

$$v_t = \sqrt{2gh}$$

$$h = \frac{v_t^2}{2g}$$

$$h = \frac{(222.22)^2}{2 \times 9.8}$$

$$h = \frac{49328.88}{19.6}$$

$$h \approx 2517.8 \, \mathrm{m}$$

So, the closest value of $$h$$ is approximately:

Option A: 2518 m.