Solution:

The point where the gravitational field is zero lies closer to the smaller mass \( m \). Let the distance of this point from \( m \) be \( x \), and from \( 4m \) be \( r - x \).

At this point:

\[
\frac{Gm}{x^2} = \frac{G \cdot 4m}{(r - x)^2}
\]

Taking the square root:

\[
\frac{1}{x} = \frac{2}{r - x}
\]

Solving:

\[
r - x = 2x \quad \Rightarrow \quad x = \frac{r}{3}
\]

The gravitational potential \( V \) at this point is the sum of the potentials due to both masses:

\[
V = -\frac{Gm}{\frac{r}{3}} - \frac{G \cdot 4m}{\frac{2r}{3}} = -\frac{3Gm}{r} - \frac{6Gm}{r} = -\frac{9Gm}{r}
\]

Thus, the potential at the point is:

\[
V = -\frac{9Gm}{r}
\]