I'll take a crack at actually answering the OP's question, or at least getting us into the ballpark.
If we replace the word 'big' with the word 'infinite', then the part of the diagram that's relevant is the rod. What we need to know is what will be the temperature drop across the length of the rod given some power (heat) input. Then the temperature at the tip of the rod is the heatsink temperature plus the drop across the rod. Conversely, we could state the maximum tip temperature, then calculate the maximum power input (probably more useful here).
The formula we need is dT = P/k * (l/A), where
dT is the temperature differential
P is the power input, watts
l is the length of the rod, meters
A is the area of the rod, m^2
k it the thermal conductivity of the material, W/m-K
IF YOU CAN GROK THIS EQUATION, YOU WILL BECOME A THERMAL GENIUS!
Note the distinction between upper and lower case: K, for Kelvin (equivalent to one °C) and k, for thermal conductivity.
For our purposes, I rearrange the formula to
P = dT*k * A/l
If we want a max tip temp of 75°C, and the heatsink temp is 25°C, then dT is 50°C or 50K
If we're using some relatively pure copper alloy, k=385 W/m-K
A= (0.01m)^2
l = 0.06m
P = 50K * 385W/m-K * 0.0001m^2 / 0.06m = 32W.
Noting that this is only the heat component, not the full electrical power input, clearly this is sufficient for almost any LED (but not large COBs). If you need a lower tip temperature, have a higher heatsink temperature, and a lower k value (and it's quite possible you could have all of these), the results could be quite different.
Knowing the acceptable tip temperature is another discussion, though extremely similar to this one. It uses the same equation several times. Grok it!