Solution:

The point where the gravitational force on the projectile is zero occurs when the gravitational forces from both planets are equal.

\[
\frac{G \cdot 4M \cdot m}{r^2} = \frac{G \cdot 9M \cdot m}{(6R - r)^2}
\]

Simplifying,

\[
\frac{4}{r^2} = \frac{9}{(6R - r)^2}
\]

Taking the square root:

\[
\frac{2}{r} = \frac{3}{6R - r}
\]

Cross-multiplying:

\[
2(6R - r) = 3r
\]

Solving:

\[
12R - 2r = 3r
\]

\[
5r = 12R
\]

\[
r = \frac{12R}{5}
\]

So, the distance from the lighter planet is \( \frac{12R}{5} \).