A

Step 1: Identify the initial conditions of the problem.
The body is initially at rest, so its initial velocity, $u = 0$. It is at a height $h = R$ above the earth’s surface.

Step 2: Use the concept of gravitational potential energy and kinetic energy conservatively. The initial potential energy at height $R$ is given by:
$$ PE_i = mgh = mgR $$
Step 3: Upon reaching the Earth's surface, the potential energy will be zero and all the initial potential energy will convert into kinetic energy (KE). The kinetic energy when the body reaches the surface is given by:
$$ KE_f = \frac{1}{2} mv^2 $$
Step 4: Setting the initial potential energy equal to the final kinetic energy:
$$ mgR = \frac{1}{2} mv^2 $$
Step 5: Cancel the mass $m$ from both sides (assuming $m \neq 0$):
$$ gR = \frac{1}{2} v^2 $$
Step 6: Multiply both sides by 2:
$$ 2gR = v^2 $$
Step 7: Take the square root to find the final velocity:
$$ v = \sqrt{2gR} $$
Step 8: Note that since we are asked for the velocity in terms of $g$, $R$, and constants, the solution will depend on these variables.
Therefore, the final velocity of the body when it reaches the Earth’s surface is:
$$ v = \sqrt{2gR} $$
This confirms the application of the conservation of energy principle in deriving the solution.