For integers $a$, $b$, $m$, write $a \equiv b \pmod{m}$ if $a - b$ is divisible by $m$. For any integer $n$ we have
$$(10n + 3)(10n + 7) = 100n^2 + 100n + 21 \equiv 21 \pmod{100}.$$
We also have
$$21^5 = (20 + 1)^5 = \sum_{k=0}^{5} \binom{5}{k} \cdot 20^k \equiv \binom{5}{1} \cdot 20^1 + \binom{5}{0} \cdot 20^0 \equiv 1 \pmod{100}.$$
Then, we see that
$$\begin{aligned} X &= (3 \times 7) \cdot (13 \times 17) \cdots (2003 \times 2007) \equiv 21^{201} \\ &\equiv (21^5)^{40} \times 21 \equiv 1^{40} \times 21 \equiv 21 \pmod{100}, \end{aligned}$$
which implies that $2$ is the desired answer.