To solve this problem, we need to find the point on the line joining the two masses where the gravitational field is zero.$\\$Consider two masses: $m$ and $9m$ separated by a distance $R. \\$Let the point where the gravitational field is zero be at a distance $x$ from mass $m. \\$Then, the distance from mass $9m$ to this point is $(R - x). \\$The gravitational field due to a mass $M$ at a distance $r$ is given by:$\\ E = \frac{GM}{r^2} \\$At the point where the gravitational field is zero, the fields due to both masses must be equal and opposite:$\\ \frac{Gm}{x^2} = \frac{G(9m)}{(R-x)^2} \\$Cancel $Gm$ from both sides:$\\ \frac{1}{x^2} = \frac{9}{(R-x)^2} \\$Taking the square root of both sides:$\\ \frac{1}{x} = \frac{3}{R-x} \\$Cross-multiplying gives:$\\ R-x = 3x \\$Solving for $x: \\ R = 4x \\ x = \frac{R}{4} \\$Thus, the point where the gravitational field is zero is $\frac{R}{4}$ from mass $m. \\$Now, we calculate the gravitational potential at this point. The gravitational potential $V$ due to a mass $M$ at a distance $r$ is given by:$\\ V = -\frac{GM}{r} \\$The total potential at the point is the sum of the potentials due to both masses:$\\ V_{total} = -\frac{Gm}{x} - \frac{G(9m)}{(R-x)} \\$Substitute $x = \frac{R}{4}$ and $R-x = \frac{3R}{4}: \\ V_{total} = -\frac{Gm}{\frac{R}{4}} - \frac{9Gm}{\frac{3R}{4}} \\$Simplify:$\\ V_{total} = -\frac{4Gm}{R} - \frac{12Gm}{R} \\ V_{total} = -\frac{16Gm}{R} \\$Therefore, the gravitational potential at the point where the gravitational field is zero is $-\frac{16Gm}{R}. \\$This corresponds to Option 3.