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

The ratio of speeds of $\text{A}$ and $\text{B}$ is given as:
$$S_A:S_B=2:3$$

Since the distance covered to reach the destination is constant, the time taken is inversely proportional to the speed ($T\propto \frac{1}{S}$). Therefore, the ratio of time taken is:

$$T_A:T_B=3:2$$

Let the time taken by $\text{A}$ be $3k$ and by $\text{B}$ be $2k$.
According to the problem, $\text{A}$ takes $15\text{ minutes}$ more than $\text{B}$:

$$T_A-T_B=15$$

$$3k-2k=15⟹k=15$$

Thus, the original time taken by $\text{A}$ is:

$$T_A=3\times 15=45\text{ minutes}$$

If $\text{A}$ doubles his speed, the time taken will be reduced to exactly half of his original time:

$$\text{New time}=\frac{T_A}{2}=\frac{45}{2}=22.5\text{ minutes}$$