1. Find the orbital speed close to Earth’s surface

For a circular orbit just above Earth’s surface,

$v_o = \sqrt{gR}$

Given:

$g = 10\ \text{m/s}^2, \qquad R = 6400\ \text{km} = 6.4 \times 10^6\ \text{m}$

So,

$v_o = \sqrt{10 \times 6.4 \times 10^6}$

$v_o = \sqrt{6.4 \times 10^7}$

$v_o = 8 \times 10^3\ \text{m/s} = 8\ \text{km/s}$

2. Find the escape speed at the same point

Escape speed is related to orbital speed by

$v_e = \sqrt{2}\, v_o$

Hence,

$v_e = \sqrt{2} \times 8 = 8\sqrt{2}\ \text{km/s}$

3. Additional speed required

The spaceship is already moving in orbit with speed $v_o$, so the extra speed needed to escape is

$\Delta v = v_e - v_o$

$\Delta v = 8\sqrt{2} - 8$

$\Delta v = 8(\sqrt{2}-1)\ \text{km/s}$

4. Match with the options

This corresponds to:

Option D: $8(\sqrt{2}-1)\ \text{km/s}$

5. Check mass dependence

The mass of the spaceship $2\times 10^4\ \text{kg}$ is irrelevant here, because orbital and escape speeds do not depend on the mass of the object.

Final Answer: $\boxed{8(\sqrt{2}-1)\ \text{km/s}}$ (Option D)