We claim that Arash has a winning strategy if and only if $k \le \frac{1400 \times 1401}{3} = 1400 \times 467$.

First of all, it is obvious if $k > 1400 \times 467$, then Arash can not color $k$ trominos in his first turn and he will lose. Indeed, there are not $k$ disjoint trominos in the table.

Next, we shall prove that for every $k \le 1400 \times 467$, Arash has a winning strategy. We can partition the table into $700 \times 467$ number of $2 \times 3$ rectangles (intersection of two consecutive columns and three consecutive rows). We count the rectangles in such a way that the rectangles filling the left columns have numbers $1, 2, \ldots, 467$ (from top to bottom), then the rectangles filling the third and fourth column have numbers $468, 469, \ldots, 934$, etc. We consider two different trominos in each $2 \times 3$ table, as in the figure below.

Now if $k \ge 700 \times 467$, Arash will color the trominos of type 1 in all the rectangles and then $k' = k - 700 \times 467$ trominos of type 2 in rectangles of numbers $1, 2, \ldots, k'$. It is easy to find that after Arash's turn, every $2 \times 2$ square has a colored cell. So, Babak can not color any $2 \times 2$ square in his turn and Arash will win.

If $k \le 700 \times 467$, Arash in his first turn colors the trominos of type 1 in rectangles $1, 2, \ldots, k$. For the next turns, suppose that Babak has filled some square and now this is Arash's turn. If he can find $k$ disjoint trominos not intersecting rectangles $1, 2, \ldots, k$, he will color them. In the case that there is not $k$ disjoint trominos, let $k'$ be the maximal number such trominos ($k' < k$). Now, he will color $k'$ trominos without intersection with rectangles $1, 2, \ldots, k$, and also color $k - k'$ trominos in rectangles $1, 2, \ldots, k - k'$. After this step, there would be no uncolored $2 \times 2$ square left in the table and so Babak will lose the game. ■