1. Condition given

The ball falls freely through air from rest through height $h$, so just before entering water its speed is

$v=\sqrt{2gh}.$

After entering water, its velocity does not change. That means its acceleration in water is zero at that instant, so the net force on the ball in water must be zero.

2. Forces on the ball inside water

When moving downward in water with speed $v$, the forces are:

• Weight downward: $W=\frac{4}{3}\pi r^3 \rho g$
• Buoyant force upward: $B=\frac{4}{3}\pi r^3 \sigma g$ where $\sigma$ is density of water.
• Viscous drag upward (Stokes' law): $F_v=6\pi \eta r v$

Since velocity remains constant after entering water,

$W=B+F_v$

So,

$\frac{4}{3}\pi r^3 (\rho-\sigma)g = 6\pi \eta r v$

Cancelling $\pi r$:

$\frac{4}{3}r^2(\rho-\sigma)g = 6\eta v$

Hence,

$v=\frac{2r^2(\rho-\sigma)g}{9\eta}$

3. Substitute values

Given:

$r=1\times 10^{-4}\,\text{m}$ $\rho=10^5\,\text{kg/m}^3$ $\sigma=10^3\,\text{kg/m}^3$ $\eta=9.8\times 10^{-6}\,\text{N s/m}^2$ $g=9.8\,\text{m/s}^2$

Now,

$v=\frac{2(10^{-4})^2(10^5-10^3)(9.8)}{9(9.8\times 10^{-6})}$

$v=\frac{2\times 10^{-8}\times 99\times 10^3\times 9.8}{9\times 9.8\times 10^{-6}}$

Cancel $9.8$:

$v=\frac{2\times 99\times 10^{-5}}{9\times 10^{-6}}$

$v=\frac{198}{9}\times 10$

$v=22\times 10=220\,\text{m/s}$

4. Use free-fall relation

Since the ball fell freely through height $h$ before entering water,

$v^2=2gh$

So,

$h=\frac{v^2}{2g} = \frac{220^2}{2\times 9.8}$

$h=\frac{48400}{19.6} \approx 2469.4\,\text{m}$

This is not exactly among the options. Let us check the usual approximation used in such problems.

If we neglect buoyancy compared to the very large density of the ball, then

$v=\frac{2r^2\rho g}{9\eta}$

$v=\frac{2(10^{-4})^2(10^5)(9.8)}{9(9.8\times 10^{-6})}$

Again cancel $9.8$:

$v=\frac{2\times 10^{-8}\times 10^5}{9\times 10^{-6}}$

$v=\frac{2\times 10^{-3}}{9\times 10^{-6}}=\frac{2}{9}\times 10^3\approx 222.2\,\text{m/s}$

Then,

$h=\frac{(222.2)^2}{19.6}\approx 2519\,\text{m}$

which matches option A.

5. Conclusion

Using the standard intended approximation $\rho \gg \sigma$, the answer is

$\boxed{2518\,\text{m}}$

So the correct option is A.