The other answers work, but are more than is necessary.  Suppose that $T$ is a finite triangulation in $R^3$.  Let $|T|$ be the _underlying space_ for $T$.  A necessary and sufficient condition for $|T|$ to be a three-ball is:

* the space $|T|$ is a manifold and
* the boundary $\partial\\,|T|$  is a two-sphere. 

These are clearly necessary.  That they suffice is a [theorem of Alexander][1], plus a bit of work.  Both conditions can be reduced to homology computations, but this is not really the "right" way to think about it.  It is more correct to think in terms of recognizing surfaces.  Namely you have to recognize all of the vertex links (each should be a sphere or a disk) and the boundary (it should be a sphere). 

Edit - here is a very brief discussion of the extra work needed.  Suppose that $|T|$ is a manifold and $S = \partial\\,|T|$ is a two-sphere.  Then by Alexander's theorem $S$ bounds a ball $B$.  You need to show that $|T|$ is equal to $B$.  This can be done using the fact that $S$ has a "collar" inside of $|T|$.  

  [1]: http://www.math.cornell.edu/~hatcher/3M/3M.pdf