I don't think the market sizing approach is correct- Here's my approach

Question

At 200m cars in the US with an average life span of 20 years, this means a replacement rate of c.10m cars a year (assuming the US car market is in steady state - which is a reasonable assumption)

70% of these are new cars with alloy wheels and the other 30% are new cars with steel wheels. So we can calculate that of the 10m new cars, 7m have alloy wheels and given that each car has 5 wheels, the number of alloy wheels sold for new cars in the US is 7m x 5 wheels. Which is 35m wheels.

We then need to calculate the number of replacement wheels sold for old cars requiring replacements

Of the 200m cars in the US, we assume 10m are condemned and replaced with new cars, and on the other hand, the average wheel life span for the rest is 10 years, meaning 190/10m = 19m cars have their wheels replaced yearly.

Similarly, 70% of these have alloy wheels, and all will have alloy wheel replacements = 70%*19*100%*5 = 66.5m wheels

Of the remaining 30% that have steel wheels, only 20% will have their wheel replaced with alloy wheels = 30%*19*20%*5 = 5.7m wheels

Total number of wheels sold in a year = 35m + 66.5m + 5.7m = 107.2m wheels

Answer 1

It is not correct indeed. Let me point out some conceptual mistakes:

Total number of sets of wheels

You state that the life span of a wheel is 10 years, and of a vehicle is 20 years. That means that the segmentation between "new" and "old" doesn't work as you assume. The "new" cars will have their alloys last for 10 years. So EVERY car that is 1-10 years old (basically 50% of the total: 100m) will be running on their first wheel and when they get to year 10 they replace the wheel. That whell will last from year 11-20 (another 50% of the total: 100m).
A simpler way to do this is to consider that each vehicle uses 2 sets of wheels over their lifetime. So over 20 years, the 200m vehicles will need 200 x 2 sets of wheels = 400 sets of wheels OVER THEIR LIFETIME (20 years).
To convert to a single year, you will need to divide by 20 years: 400m / 20 years = 20m sets per year.
Let me note this: I would rather have you divide the population in two segments: cars year 1-10; and cars year 11-20. Why? Check the next part of the answer.

Total number of wheels in a set

You assume that you'll need 5 wheels per set. This is wrong. You need 5 wheels for the new vehicles (years 1-10). But when they get to year 10 they need to replace 4 wheels, but not the spare wheel. So you have to assume different values here.

A few more issues:

You are assuming that penetration of alloy wheels will increase for older cars. I believe one could argue the opposite. When the car is older you don't spend money to improve the car, you rather consider that it is approaching the end of it's useful life, you are probably making fewer miles with that car (more likely to be a second car) and would rather use worse parts. So instead of INCREASING penetration, I would decrease.
You are ignoring that accidents may require changing a wheel before it reaches its useful life.
You are ignoring that US population is increasing, which means that assuming that there's a stable vehicle park is quite a stretch... this would mean a decrease in vehicles per capita. One could argue that is the case, but this point would have to be verbalized and justified. There are also some other effects could be considered (e.g. exports of used vehicles).
Please note that I would not expect a candidate to consider both of these (increasing population, exports) but at least having one of them (or at least justifying Why it doesn't have a meaningful impact) would be expected.