1. Set up the configuration

Let the large fixed sphere of mass $M$ be at the origin.

The two identical point masses $m$ lie on the same line, connected by a rigid massless rod of length $l$.

When the nearer mass is at distance $3l$ from $M$, the farther mass is at distance $3l+l=4l.$

So:

• inner mass: at $r_1=3l$
• outer mass: at $r_2=4l$

2. Meaning of zero tension

If the tension in the rod is zero, then the rod is just not required to push/pull either mass. That means both masses must naturally have the same acceleration under gravity alone, so that their separation remains constant without any internal force.

Thus, $a_1=a_2.$

3. Acceleration of the inner mass

The inner mass experiences:

• attraction by $M$ toward the origin: $F_{M\to 1}=\frac{GMm}{(3l)^2}=\frac{GMm}{9l^2}$
• attraction by the outer mass $m$ toward the outer side (opposite direction), since the outer mass is at distance $l$ from it: $F_{2\to 1}=\frac{Gm^2}{l^2}$

Taking inward (toward $M$) as positive, the net force on inner mass is $F_1=\frac{GMm}{9l^2}-\frac{Gm^2}{l^2}.$

Hence its acceleration is $a_1=\frac{F_1}{m}=\frac{GM}{9l^2}-\frac{Gm}{l^2}.$

4. Acceleration of the outer mass

The outer mass experiences:

• attraction by $M$ inward: $F_{M\to 2}=\frac{GMm}{(4l)^2}=\frac{GMm}{16l^2}$
• attraction by the inner mass also inward: $F_{1\to 2}=\frac{Gm^2}{l^2}$

So net inward force is $F_2=\frac{GMm}{16l^2}+\frac{Gm^2}{l^2}.$

Hence acceleration is $a_2=\frac{F_2}{m}=\frac{GM}{16l^2}+\frac{Gm}{l^2}.$

5. Equate accelerations for zero tension

$\frac{GM}{9l^2}-\frac{Gm}{l^2}=\frac{GM}{16l^2}+\frac{Gm}{l^2}.$

Multiply by $\dfrac{l^2}{G}$: $\frac{M}{9}-m=\frac{M}{16}+m.$

So, $\frac{M}{9}-\frac{M}{16}=2m.$

$M\left(\frac{16-9}{144}\right)=2m$

$\frac{7M}{144}=2m$

$m=\frac{7M}{288}.$

6. Compare with the given form

Given $m=k\left(\frac{M}{288}\right).$

Comparing with $m=\frac{7M}{288},$ we get $k=7.$

7. Final answer

$\boxed{7}$