The value of acceleration due to gravity at Earth's surface is 9.8 m s$$^{-2}$$. The altitude above its surface at which the acceleration due to gravity decreases to 4.9 m s$$^{-2}$$, is close to: (Radius of earth = $$6.4 \times 10^6$$ m)

We know that the acceleration due to gravity at any distance from the centre of Earth is given by the Newtonian formula $$g \;=\; \dfrac{G\,M}{r^{\,2}}$$ where $$G$$ is the gravitational constant, $$M$$ is the mass of Earth and $$r$$ is the distance of the point from the centre of Earth.

At Earth’s surface the distance from the centre is simply the radius $$R$$ of Earth, so the surface value is

$$g \;=\; \dfrac{G\,M}{R^{\,2}} \;=\; 9.8 \text{ m s}^{-2}.$$

If we go to an altitude $$h$$ above the surface, the distance from the centre becomes $$R+h$$ and the acceleration due to gravity there (call it $$g'$$) is

$$g' \;=\; \dfrac{G\,M}{(R+h)^{2}}.$$

To find the relation between $$g'$$ and $$g$$, we divide the two expressions:

$$\dfrac{g'}{g} \;=\; \dfrac{\dfrac{G\,M}{(R+h)^{2}}}{\dfrac{G\,M}{R^{\,2}}} \;=\; \dfrac{R^{\,2}}{(R+h)^{2}}.$$

Simplifying the right-hand side, we get the useful relation

$$g' \;=\; g\;\left(\dfrac{R}{R+h}\right)^{2}.$$

According to the question, the value at altitude must be one-half of the surface value, i.e.

$$g' = 4.9 \text{ m s}^{-2} = \dfrac{1}{2}\,g.$$

Substituting this condition into the previous relation, we have

$$\dfrac{1}{2}\,g \;=\; g\;\left(\dfrac{R}{R+h}\right)^{2}.$$

Now we cancel the common factor $$g$$ on both sides:

$$\dfrac{1}{2} \;=\; \left(\dfrac{R}{R+h}\right)^{2}.$$

Taking the square root of both sides (and keeping only the positive root because distances are positive), we obtain

$$\sqrt{\dfrac{1}{2}} \;=\; \dfrac{R}{R+h},$$

so

$$\dfrac{1}{\sqrt{2}} \;=\; \dfrac{R}{R+h}.$$

Cross-multiplying gives

$$R \;=\; \dfrac{R+h}{\sqrt{2}}.$$

Rewriting this in a clearer form, multiply both sides by $$\sqrt{2}$$:

$$\sqrt{2}\,R \;=\; R + h.$$

Now we isolate $$h$$ by subtracting $$R$$ from both sides:

$$h \;=\; (\sqrt{2} - 1)\,R.$$

We substitute the numerical value of Earth’s radius $$R = 6.4 \times 10^{6} \text{ m}$$ and the approximate value $$\sqrt{2} \approx 1.414$$:

$$h \;=\; (1.414 - 1)\,(6.4 \times 10^{6}) \text{ m}.$$

This difference $$1.414 - 1 = 0.414$$, so

$$h \;=\; 0.414 \times 6.4 \times 10^{6} \text{ m}.$$

Multiplying $$0.414$$ and $$6.4$$ step by step:

$$0.414 \times 6 = 2.484,$$

$$0.414 \times 0.4 = 0.1656,$$

Adding, $$2.484 + 0.1656 = 2.6496.$$

Thus

$$h \approx 2.65 \times 10^{6} \text{ m}.$$

Among the given options the closest numerical value is $$2.6 \times 10^{6} \text{ m}$$, which corresponds to Option B.

Hence, the correct answer is Option B.