Let the point where gravitational field is zero be between the masses.

Place mass $m$ at one end and mass $4m$ at the other, separated by distance $r$.

Let the required point be at distance $x$ from mass $m$. Then its distance from mass $4m$ is $(r-x)$.

Condition for zero gravitational field

The gravitational fields due to the two masses at that point must be equal in magnitude and opposite in direction:

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

Cancelling $Gm$:

$\frac{1}{x^2}=\frac{4}{(r-x)^2}$

$(r-x)^2 = 4x^2$

Taking positive root (since distances are positive):

$r-x=2x$

$r=3x$

$x=\frac{r}{3}$

So the point is at distance $\frac{r}{3}$ from mass $m$ and $\frac{2r}{3}$ from mass $4m$.

Gravitational potential at that point

Gravitational potential is the scalar sum:

$V=-\frac{Gm}{x}-\frac{G(4m)}{r-x}$

Substitute $x=\frac{r}{3}$ and $r-x=\frac{2r}{3}$:

$V=-\frac{Gm}{r/3}-\frac{4Gm}{2r/3}$

$V=-\frac{3Gm}{r}-\frac{6Gm}{r}$

$V=-\frac{9Gm}{r}$

Match with options

$V=-\frac{9Gm}{r}$

This corresponds to Option C.