C

Step 1: We understand that the acceleration due to gravity at a height above the Earth's surface is defined by the formula:
$ g_h = \frac{g}{(1 + \frac{h}{R})^2} $
where $g$ is the acceleration due to gravity at the surface, $R$ is the radius of the Earth, and $h$ is the height above the surface. When $h = \frac{R}{20}$ (given), we have:

• $ g_h = \frac{g}{(1 + \frac{1}{20})^2} = \frac{g}{(1.05)^2} $
• $ g_h = \frac{g}{1.1025} $

Step 2: Consider the point at an equal distance below the surface. At a distance $d$ below the surface, the acceleration due to gravity $g_d$ is given by:
$ g_d = g \left(1 - \frac{d}{R}\right) $
where $d = \frac{R}{20}$. Hence, substituting:

• $ g_d = g \left(1 - \frac{1}{20}\right) = g \left(\frac{19}{20}\right) $

Step 3: Utilizing the fact that at a height of $\frac{R}{20}$ the $g$ is 9 m/s², we have:

• $ 9 = g \times \frac{1}{1.1025} $ to find $g$:
• $ g = 9 \times 1.1025 \approx 9.92 \, \text{m/s}^2 $

Step 4: Now substituting for $g$ in the formula for $g_d$:

• $ g_d = 9.92 \times \frac{19}{20} = 9.92 \times 0.95 \approx 9.42 \, \text{m/s}^2 $

Conclusion: Therefore, the approximate value of acceleration due to gravity at that point below the surface is roughly 9.42 m/s², which rounds to option C.