1. Use the relation for gravitational acceleration at the surface of Earth

   $g = \frac{GM}{R^2}$

   where:

   • $G$ = gravitational constant
   • $M$ = mass of Earth
   • $R$ = radius of Earth

2. Write mass in terms of average density

   Average density of Earth is

   $\rho = \frac{M}{\frac{4}{3}\pi R^3}$

   So,

   $M = \rho \cdot \frac{4}{3}\pi R^3$

3. Substitute into the expression for $g$

   $g = \frac{G\left(\rho \cdot \frac{4}{3}\pi R^3\right)}{R^2}$

   Simplifying,

   $g = \frac{4}{3}\pi G \rho R$

4. Express density in terms of $g$

   $\rho = \frac{3g}{4\pi G R}$

5. Interpret the result

   Since $G$ and $R$ are constants for Earth,

   $\rho \propto g$

   Therefore, the average density of the Earth is directly proportional to $g$.

6. Check options

   • A: False
   • B: False
   • C: False
   • D: True

Hence, the correct option is D.