The trace is a linear map from $V$ onto $\mathbb{R}$; that means that $U$ has dimension $8$, since it is equal to the kernel of this map.
The orthogonal complement will then be spanned by a single vector. You can find such a vector by finding one which is orthogonal to each vector in a basis for $U$. For example, you can take as your basis $E_{12}$, $E_{13}$, $E_{21}$, $E_{23}$, $E_{31}$, $E_{32}$, $E_{11}-E_{22}$, and $E_{11}-E_{13}$. This will give you information about $w$.
For instance, the fact that $\langle E_{12},w\rangle = 0$ tells you that the $(1,2)$ entry of $w$ must be equal to $0$. Using the rest of the basis, you should get all the information you need about the structure of $w$.
Or you could realize that there is a very special set of matrices $M$ such that for all $A$, $\mathrm{trace}(A)=0$ if and only if $\mathrm{trace}(M^tA) = 0$, and go from there. Hint. Do you know some family of matrices for which it is easy to compute the trace of $M^tA$ in terms of the trace of $A$?