Obviously $m > 9$. If $m = \overline{ab}$, where $a \geq 1$ then we must have $b = 1$ to ensure last digit of $1 \cdot m$ is equal $1$. In this case regardless of value of $a$ the second last digit of $\underbrace{11 \ldots 1}_{k} \cdot m$ is equal to the last digit of $a + 1$ and can't be equal $1$. So $m \geq 100$. Obviously $m = 100$ doesn't satisfy to the condition, but $m = 101$ satisfies, since $11 \cdot 101 = 1111$.
Hence, the answer to this problem is $m = 101$. $\square$