Is the problem with the phrase "on average"?

Perhaps. You might also be jumping to conclusions about what's being asked. The question says:

After the 100th step, how far away am I, on average, from where I started?

I'm sure you know that to calculate an average, you add all the outcomes and divide the total by the number of outcomes. Since the coin is fair, your intuition might tell you that since heads and tails occur with equal frequency, they should cancel each other out and the average distance from the starting point will be zero. However, if you think about all the possible outcomes that you could get from tossing the coin 100 times, the only ones that result in a distance of 0 steps from the start are the ones where there are exactly 50 heads and 50 tails. All the outcomes with more heads than tails or vice versa put you some distance greater than 0 steps from the start, so the average distance has to be more than 0 steps.

Here's a table with all the possible outcomes for just 4 coin tosses, along with their average distance from the start:

As you can see, if you toss a coin 4 times and take 1 step north on heads and 1 step south on tails, on average you'll end up 1.5 steps from the start.

Is the problem with the phrase "on average"?

Perhaps. You might also be jumping to conclusions about what's being asked. The question says:

After the 100th step, how far away am I, on average, from where I started?

I'm sure you know that to calculate an average, you add all the samples and divide the total by the number of samples. Since the coin is fair, your intuition might tell you that since heads and tails occur with equal frequency, they should cancel each other out and the average distance from the starting point will be zero. However, if you think about all the possible outcomes that you could get from tossing the coin 100 times, the only ones that result in a distance of 0 steps from the start are the ones where there are exactly 50 heads and 50 tails. All the outcomes with more heads than tails or vice versa put you some distance greater than 0 steps from the start, so the average distance has to be more than 0 steps.

Here's a table with all the possible outcomes for just 4 coin tosses, along with their average distance from the start:

As you can see, if you toss a coin 4 times and take 1 step north on heads and 1 step south on tails, on average you'll end up 1.5 steps from the start.