Let's break down the problem:

A takes 2 hours more than B to walk d km: So, A's time = B's time + 2 hours

If A doubles his speed, he can make it in 1 hour less than B: Let's say A's original speed is "s" km/h. So, when A doubles his speed, it becomes "2s" km/h. Now, the time it takes for A to walk d km at the doubled speed is d/(2s) hours. According to the problem, this time is 1 hour less than B's time: d/(2s) = B's time - 1

Now, we can set up a system of equations to solve for B's time and A's speed:

From the first statement: A's time = B's time + 2 From the second statement: d/(2s) = B's time - 1

Now, let's solve these equations simultaneously:

A's time = B's time + 2 d/(2s) = B's time - 1

Substitute A's time from the first equation into the second equation:

d/(2s) = (A's time - 2) - 1

Now, we know A's time is 2 hours more than B's time, so we can replace A's time with B's time + 2:

d/(2s) = (B's time + 2 - 2) - 1

Simplify:

d/(2s) = B's time - 1

Now, we have an equation relating B's time and A's speed:

d/(2s) = B's time - 1

We also know that A's time is 2 hours more than B's time:

A's time = B's time + 2

Now, let's solve for B's time:

B's time = d/(2s) + 1

So, B's time to walk d km is d/(2s) + 1 hour.

Now, we need to find A's speed "s" to calculate B's time. We can use the fact that A takes 2 hours more than B to walk d km:

A's time = B's time + 2

d/s = (d/(2s) + 1) + 2

Now, let's solve for s:

d/s = d/(2s) + 3

Multiply both sides by "s" to get rid of the fractions:

d = d/2 + 3s

Now, isolate 3s on one side:

3s = d - d/2

Combine the fractions on the right side:

3s = (2d - d)/2

3s = d/2

Now, solve for s:

s = (d/2) / 3

s = d/6

So, A's speed is d/6 km/h.

Now that we have A's speed, we can calculate B's time:

B's time = d/(2s) + 1

B's time = d / [2 * (d/6)] + 1

B's time = d / (d/3) + 1

B's time = 3 + 1

B's time = 4 hours

Therefore, B requires 4 hours to walk d km.