Take a vector $v$ not in $\ker(A - 3 I)$. Compute $u=(A - 3I)v$.
Complete $u$ to a basis of $\ker(A - 3 I)$ with some $w$.
A basis that works is $(u,v,w)$.
The reason this works is as follows. Given that you know that that $\ker(A - 3 I)^2$ is the full space, you know that the maximal size of Jordan block is $2$.
And, since $\ker(A - 3 I)$ is not the full space you know that there is a Jordan block of size greater $1$.
Thus the only way this can happen is you have one block of size $2$ and one block of size $1$.
For the Jordan block of size $2$ you need a vector $v$ not in $\ker(A - 3 I)$ itself and its image $u=(A - 3 I)v$. Note that $u = (A- 3I)v$ means that $Av= u + 3v$. Compare this to the the second colon in the Jordan block where you just have $1$ and $3$.
So with those two you have the Jordan block of size two covered. Then you need another vector in the kernel to get another block of size one.