What is the geometry of an alkyl radical?
First off, it was very astute of you to recognize that if the radical formed at carbon 3 (the methylene carbon in the starting compound) is pyramidal, then the radical would be chiral ( the lone radical electron serving as the 4th different substituent on that carbon) and 4 different radicals would be possible.

In a typical carbocation, the carbon bearing the positive charge is $\ce{sp^2}$ hybridized, meaning that the p-orbital at the cationic center is empty and the carbocation is planar.  On the other hand, in a typical carbanion, the negatively charged carbon is pyramidal and approaches $\ce{sp^3}$ hybridization. This $\ce{sp^3}$-like orbital contains 2 electrons.  Therefore, it would be reasonable to expect that a typical hydrocarbon radical, where the orbital holds 1 electron would be somewhere in between these two extremes and be slightly pyramidal.  



This is indeed the case and the degree of non-planarity is very sensitive to the substituents attached to the radical center.  In the case of hydrocarbon radicals, such as those created from your starting material, the barrier to inversion from one pyramidal form to the other is extremely low, on the order of a few kcal/mole or so.  Therefore, interconversion between the two enantiomeric forms of the radical would be extremely rapid at room temperature.  

So yes, the $\ce{CH_3\dot{C}HC(CH3)3}$ radical is slightly pyramidal and it does transition rapidly between the two pyramidal enantiomers.  Your answer of 4 radicals is correct, as long as you understand that 2 of them rapidly interconvert at room temperature.