Well define $\omega,\omega_1, \omega_2$ to be the first three infinite ordinals. Order them according to their size:
- $2\cdot\omega_1+\omega\cdot3+3,$
- $\omega\cdot3+\omega_1+3,$
- $\omega_1^{\omega_2}+\omega_2+\omega_1,$
- $\omega_1\cdot \omega_2 $
I think 3 is the biggest, 4 is equal to $\omega_2$ so that's second largest, third largest is probably 1 because the there are two limit ordinals in succession.
How do you treat the ordinal in the exponent ?
The answer to this question depends quite heavily on the answer to the following two questions:
Ordinal and cardinal arithmetic agree on finite ordinals, but otherwise may differ greatly. As for the second question, there are (for example) uncountably many ordinals having the same cardinality as $\omega.$
Let's see how the answers will differ in each case. I will denote cardinal addition, multiplication, and exponentiation by $\alpha\oplus\beta,\alpha\odot\beta,$ and ${}^\beta\alpha,$ respectively, to distinguish from the standard notation (which I will use for ordinal arithmetic).
It's worth noting that cardinal arithmetic makes things simpler in some ways and more frustrating in others. In particular, given non-$0$ initial ordinals (that is, natural numbers or ordinals $\omega_\alpha$) $\beta,\gamma,$ if at least one of $\beta,\gamma$ is infinite, then $\beta\oplus\gamma=\beta\odot\gamma=\max\{\beta,\gamma\}.$ That's fairly nice. However, without the Axiom of Choice, it turns out that cardinal exponentiation fails to be an operation on the initial ordinals--that is, for sufficiently large initial ordinals $\gamma$, we have that ${}^\gamma\beta$ fails to be an ordinal whenever $\beta$ is an initial ordinal with $2\le\beta\le\gamma.$ Let's put such concerns aside, though, and simply note that if $\gamma$ is an infinite initial ordinal, and $\beta$ is an initial ordinal with $2\le\beta\le\gamma,$ then ${}^\gamma\beta={}^\gamma2,$ whatever that may mean.
It's also worth noting that ordinal arithmetic on initial ordinals has certain properties known as indecomposability properties. In particular:
There are other ordinals with these indecomposability properties (such as the ordinals called epsilon numbers), but we're only concerned with the initial ordinals, here.
Now, let's get down to business.
Using cardinal arithmetic is simple enough: $$2\odot\omega_1\oplus\omega\odot 3\oplus 3=\omega_1\oplus\omega\oplus 3=\max\{\omega_1,\omega,3\}=\omega_1\\\omega\odot 3\oplus\omega_1\oplus 3=\omega\oplus\omega_1\oplus 3=\max\{\omega,\omega_1,3\}=\omega_1\\{}^{\omega_2}\omega_1\oplus\omega_2\oplus\omega_1={}^{\omega_2}2\oplus\omega_2\oplus\omega_1=\max\left\{{}^{\omega_2}2,\omega_2,\omega_1\right\}={}^{\omega_2}2\\\omega_1\odot\omega_2=\max\{\omega_1,\omega_2\}=\omega_2$$
Hence, in terms of cardinality, the third is the largest, followed by the fourth, followed by the first two. Without sufficient Choice, though, the third need not be an initial ordinal at all, and so sorting them that way may no longer make sense. With sufficient choice, the sorting in terms of order type will be the same as that in terms of cardinality.
The ordinal arithmetic is a bit less simple until indecomposability properties (and left-distributivity) are observed: $$2\cdot\omega_1+\omega\cdot 3+3=\omega_1+\omega\cdot 3+3\\\omega\cdot 3+\omega_1+3=\omega\cdot 3+\omega\cdot\omega_1+3=\omega\cdot(3+\omega_1)+3=\omega\cdot\omega_1+3=\omega_1+3\\\omega_1^{\omega_2}+\omega_2+\omega_1=\omega_2+\omega_2+\omega_1=\omega_2\cdot 2+\omega_1\\\omega_1\cdot\omega_2=\omega_2$$ Here, the third and fourth have the same cardinality, which is greater than the (shared) cardinality of the first and second. In terms of order type, from greatest to least, it goes: third, fourth, first, second.
My suspicion is that it is supposed to be ordinal arithmetic and that "size" is intended to mean order type, but I can't say for sure. If I used any results that are unfamiliar to you, it would be a good exercise to try to prove them. Please feel free to ask for clarification if you need it.