The ratio between the speed of travelling of A and B is $2:3$ and therefore A takes $15$ minutes more than the time taken by B to reach a destination. If A had walked at double the speed, how long would he have taken to cover the distance?

Let the speed of A be $2s$ and the speed of B be $3s$.
Since speed and time are inversely proportional when distance is constant ($\text{Time}\propto \frac{1}{\text{Speed}}$), the ratio of time taken by A and B to cover the same distance will be:

$$\text{Ratio of time (A : B)}=3:2$$

Let the time taken by A be $3t$ and by B be $2t$.
According to the question, A takes $15$ minutes more than B:

$$3t-2t=15\text{ minutes}$$

$$t=15\text{ minutes}$$

Therefore, the original time taken by A at his normal speed is:

$$\text{Original time taken by A}=3t=3\times 15=45\text{ minutes}$$

The question asks for the time taken if A travels at double his speed. Since doubling the speed cuts the time taken in half ($\text{Time}=\frac{\text{Original Time}}{2}$):

$$\text{New time taken by A}=\frac{45}{2}=22.5\text{ minutes}$$