Where will halogenation take place?


> 
> In the following halogenation reaction:
>  [![enter image description here](https://i.stack.imgur.com/7yrEo.png)](https://i.stack.imgur.com/7yrEo.png)
> 
> 
> **At which position will halogenation take place?**
> 
> 
> 


It is obvious that the halogenation will take place at a 3° Carbon atom (the rates of halogenation for 3°,2°,1° atoms follow the ratio 1600:82:1 or thereabouts). **The question is which 3° atom?** I labeled the atoms as shown:


[![enter image description here](https://i.stack.imgur.com/J8Y4U.png)](https://i.stack.imgur.com/J8Y4U.png)


I believe that halogenation will take place at **A** because the cyclopropyl ring can stabilize the free radical formed at A better than the 6-membered ring. A similar question that deals with carbocations is [here](https://chemistry.stackexchange.com/questions/19090/what-is-the-reason-for-the-exceptional-stability-of-the-cyclopropylmethyl-carboc), which shows the stability of the cyclopropylmethyl carbocation. Can a similar logic also be applied to free radicals? Please confirm.


