3. A student forgot to write the multiplication sign between two three-digit numbers, resulting in a six-digit number. This six-digit number is exactly three times the product of the original two three-digit numbers. Then, this six-digit number is $\qquad$
3. 167334 .

Let the two three-digit numbers be $x$ and $y$.
From the problem, we have
$$
\begin{array}{l}
1000 x + y = 3 x y \\
\Rightarrow y = 3 x y - 1000 x = (3 y - 1000) x .
\end{array}
$$

Thus, $y$ is an integer multiple of $x$, let's assume $y = t x$ ($t$ is a positive integer). Substituting into equation (1) we get
$$
1000 + t = 3 t x \Rightarrow x = \frac{1000 + t}{3 t} \text {. }
$$

Since $x$ is a three-digit number, we have
$$
x = \frac{1000 + t}{3 t} \geqslant 100 \Rightarrow t \leqslant \frac{1000}{299} \text {. }
$$

Since $t$ is a positive integer, the possible values of $t$ can only be $1, 2, 3$.
Upon verification, only $t = 2$ satisfies the condition.

Thus, $t = 2, x = 167, y = 334$.
Therefore, the required six-digit number is 167334.