1. Weight at height $h$ above Earth's surface

   The gravitational acceleration at a distance $R+h$ from the center of Earth is $g_h = g\left(\frac{R}{R+h}\right)^2$ Hence the weight there is proportional to $W_h = mg\left(\frac{R}{R+h}\right)^2$

2. Weight at depth $h$ below Earth's surface

   Inside Earth, assuming uniform density, gravitational acceleration varies linearly with distance from the center: $g_d = g\left(1-\frac{h}{R}\right)$ Hence the weight at depth $h$ is $W_d = mg\left(1-\frac{h}{R}\right)$

3. Given condition: weights are equal

   $mg\left(\frac{R}{R+h}\right)^2 = mg\left(1-\frac{h}{R}\right)$

   Cancel $mg$: $\left(\frac{R}{R+h}\right)^2 = 1-\frac{h}{R}$

4. Let $x = \frac{h}{R}$ Then the equation becomes $\frac{1}{(1+x)^2} = 1-x$

5. Solve the equation

   $1 = (1-x)(1+x)^2$

   Expand: $1 = (1-x)(1+2x+x^2)$ $1 = 1 + 2x + x^2 - x - 2x^2 - x^3$ $1 = 1 + x - x^2 - x^3$

   Rearranging: $x - x^2 - x^3 = 0$ $x(1 - x - x^2)=0$

   Since $h \neq 0$, we take $1 - x - x^2 = 0$ $x^2 + x - 1 = 0$

   $x = \frac{-1 \pm \sqrt{1+4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$

   Only the positive root is physical: $x = \frac{\sqrt{5}-1}{2}$

   Therefore, $h = xR = \frac{\sqrt{5}-1}{2}R = \frac{\sqrt{5}R - R}{2}$

6. Match with options

   This corresponds to Option B.

Final Answer: $\boxed{\frac{\sqrt{5}R - R}{2}}$