Answer: 2.

Solution: Let $x$ be the time it takes for Dima to walk from home to school, $y$ be the time it takes for Dima to run from home to school, and $T$ be the remaining time until the bell rings (at the moment Dima noticed the loss). Then the conditions of the problem can be written as $\left\{\begin{array}{l}\frac{x}{2}=T-3 \\ \frac{y}{2}+y=T+3 \\ \frac{y}{2}+x=T+15\end{array}\right.$ Solving this, we get that $x=24, y=12, T=15$, from which it follows that he runs twice as fast as he walks.
6. Dima went to school in the morning, but after walking exactly half the distance, he realized he had forgotten his mobile phone at home. Dima estimated (he had an A in mental arithmetic) that if he continued walking at the same speed, he would arrive at school 3 minutes before the first bell, but if he ran home for the phone and then ran to school, he would arrive 3 minutes after the bell. Dima decided to run home, but he got out of breath while running (he had a C in physical education), and walked from home to school at his usual speed. As a result, he was 15 minutes late for the first class! How many times faster is the speed at which he runs compared to the speed at which he walks?