The breaking stress of a wire, denoted as $$\sigma$$, is the maximum tension per unit area it can withstand before it breaks. It is given as:

$ \sigma = \frac{F}{A} $

where $F$ is the breaking force and $A$ is the cross-sectional area of the wire. The weight of the wire, when it's on the verge of breaking, equals the maximum force $F$ it can sustain. The weight of an object is given by:

$ W = mg $

where $m$ is the mass of the object and $g$ is the acceleration due to gravity. Since the wire's mass, $m$, can also be expressed in terms of its density ($\rho$), length ($L$), and area ($A$) as:

$ m = \rho V = \rho A L $

We can substitute this expression in the equation for weight:

$ W = \rho A L g $

On a planet where the acceleration due to gravity is $\frac{1}{3}$rd of that on Earth, we substitute $g_{\text{planet}} = \frac{1}{3}g_{\text{Earth}} = \frac{1}{3} \times 10 \, \text{m/s}^2 = \frac{10}{3} \, \text{m/s}^2$. The breaking force $F$ which is equivalent to the weight at the breaking point, is therefore given by:

$ F = \rho A L g_{\text{planet}} $

Since the breaking stress $\sigma$ is also $F/A$, we can set the two expressions equal to find the maximum length $L$ of the wire before breaking:

$ \sigma = \frac{\rho A L g_{\text{planet}}}{A} $

$ \sigma = \rho L g_{\text{planet}} $

Solving for $L$:

$ L = \frac{\sigma}{\rho g_{\text{planet}}} $

Given that $\sigma = 1.2 \times 10^8 \, \text{N/m}^2$, $\rho = 6 \times 10^4 \, \text{kg/m}^3$, and $g_{\text{planet}} = \frac{10}{3} \, \text{m/s}^2$, we can substitute these values into the formula:

$ L = \frac{1.2 \times 10^8}{6 \times 10^4 \times \left( \frac{10}{3} \right) } $

$ L = \frac{1.2 \times 10^8}{2 \times 10^5} $

$ L = 600 \, \text{m} $

Therefore, the maximum length of the wire that can be suspended from a rigid support on this planet without breaking is $600$ meters.