A more conceptual proof:
Le $\mathcal B$ a basis of $V$. By definition of a basis, we have an isomorphism $L(V,W)\simeq W^\mathcal B$, so $$\dim(L(V,W))=\dim(W^\mathcal B)=\lvert\mathcal B\rvert\cdot \dim W =\dim V\cdot \dim W.$$
Your reasoning seems good, but you have not written the description of $T$. Therefore we can't say if your answer is right or wrong.
But let's go through what you have written. You have talked about a basis of $V $, namely $u_1,u_2,u_3,v_1,v_2$, out of which $u_i$ are in the null space. This directly translates to $T(u_i)=0$.
Now, all you need to do is ensure that $T(v_1)$ and $T(v_2)$ are linearly independent. The best way to do this is to simple consider two linearly independent vectors in $W$, say $w_1$ and $w_2$ (if $\dim W < 2$, you can't construct such a $T$!), and map $v_1 \to w_1$ and $v_2 \to w_2$.
Therefore, your full definition of the operator $T$ is like this: Let $x \in V$ be of the form $x = x_1u_1 + x_2u_2 + x_3u_3 + x_4v_1+x_5v_2$. Then, define $T: V \to W$ by $T(x) = x_4w_1 + x_5w_2$.
I leave you to check that $\dim [\text{null }T] = 3$ and $\dim [\text{range }T] = 2$.
Just a friendly reminder : If $\dim W < 2$, then no such $T$ exists!
To give an explicit example : define $T : \mathbb R^5 \to \mathbb R^5$ by $T(a,b,c,d,e) = (a,b,0,0,0)$. This has rank two, nullity three.
Best Answer
Hint: define $T$ by $v_i\longmapsto w_i$, $i=1,\dots n$.