Solution:

The average density of the Earth (\( \rho \)) is directly proportional to the acceleration due to gravity (\( g \)) at the surface due to the relationship expressed by the formula:

\[
g = \frac{G M}{R^2},
\]

where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth.

The mass \( M \) can be expressed in terms of density and volume:

\[
M = \rho V = \rho \left( \frac{4}{3} \pi R^3 \right).
\]

Substituting this into the equation for \( g \):

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

This shows that \( g \) is directly proportional to \( \rho \) (as \( R \) and \( G \) are constants), meaning that as the average density of the Earth increases, the value of \( g \) also increases.