1. Total energy of a satellite in a circular orbit

For a satellite of mass $m$ in a circular orbit of radius $r$, the total mechanical energy is

$E = -\frac{GMm}{2r}$

Using $GM = gR^2$, this becomes

$E = -\frac{gR^2 m}{2r}$

2. Initial orbit

Given:

• $m = 10^3\,\text{kg}$
• initial radius $r_1 = 2R$
• $g = 10\,\text{m/s}^2$

So initial total energy is

$E_1 = -\frac{gR^2 m}{2(2R)} = -\frac{gRm}{4}$

Substitute values:

$E_1 = -\frac{10 \cdot R \cdot 10^3}{4} = -2500R\,\text{J}$

3. Energy supplied

Given supplied energy:

$\Delta E = \frac{10^4 R}{6}\,\text{J}$

This increases the total energy of the satellite.

Hence new total energy:

$E_2 = E_1 + \Delta E$

$E_2 = -2500R + \frac{10^4R}{6}$

Take LCM:

$-2500R = -\frac{15000R}{6}$

So,

$E_2 = -\frac{15000R}{6} + \frac{10000R}{6} = -\frac{5000R}{6} = -\frac{2500R}{3}$

4. New circular orbit radius

For new orbit radius $r_2$,

$E_2 = -\frac{gR^2 m}{2r_2}$

Substitute $g=10$, $m=10^3$:

$E_2 = -\frac{10\cdot R^2 \cdot 10^3}{2r_2} = -\frac{5000R^2}{r_2}$

Set this equal to the value found:

$-\frac{5000R^2}{r_2} = -\frac{2500R}{3}$

Cancel minus sign:

$\frac{5000R^2}{r_2} = \frac{2500R}{3}$

Multiply both sides by $r_2$:

$5000R^2 = \frac{2500R}{3} r_2$

$r_2 = \frac{5000R^2 \cdot 3}{2500R} = 6R$

5. Option check

The new circular orbit radius is

$\boxed{6R}$

So the correct option is B.