220 is 2^2 * 5 * 11, it can be obtained as (2,10,11) and (4,5,11)
216 is 2^3 * 3^3, it can be obtained as (2,9,12), (3,6,12), (3,8,9), (4,6,9), (2,3,4,9)
224 is 2^5 * 7, it can be obtained as (14,16), (2,7,16), (2,8,14), (4,7,8)
Then, next closest is 225 (3,5,15), followed by 210 - (2,3,5,7), (2,7,15), (5,6,7), (3,7,10).
Sometimes solution refers to "we" = first player, and "he" = second player. Sorry for not sticking to any particular way.
WARNING, SPOILERS BELOW!
As we can see, 220 is trivial to block. 216 and 224 are slightly harder to block with more combinations. Out of these numbers, 2 and 4 pop out - they are both present in both the 220 and 216/224 combos. So, it seems the winning strategy should pick one of those two. Let's first try with 2. Second player then picks
Anything but 4: we pick 4. Second player cannot assemble 220, and picks second number:
0) He doesn't have any of 5,10,11 - we pick 11, then 5 or 10 in the next turn assembling 220, we win.
So, he has one of these three numbers plus:
1A) Anything but 9, 14 or 16 - first player picks 9, then (4,6,9), (2,3,4,9) or (2,9,12) is inevitable. Blocking any of second player attempts at his own 216/224 is trivial and costless.
1B) On picking 14 or 16, first player grabs the other number, then the second player is required to block one of the (2,8,14) or (2,7,16). First player then picks 9. 216 is forced.
1C) On picking 9, we grab 14. Either (2,8,14) or (14,16) cannot be blocked.
So, now for the interesting case where the second player picked 4. We ignore the 220 possibility and NOT pick anything that would lead towards it (if we pick 11 he gets 10, that's it)
What are the options left? Well, both can still get (14,16), (3,6,12) and (3,8,9). The first player can hope for (2,9,12), (2,7,16) and (2,8,14). The second can hope for (4,6,9) and (4,7,8). (2,3,4,9) is gone.
Here, the immediate threat of the second player is getting 8 and 9, opening triplets (3,8,9), (4,6,9) and (4,7,8). If we pick something outside of these three sets (eg 14), he gets 9, so we need 6, then he gets 8, so we need 7, then he gets 3 and draws. So, we need to get one of these numbers. Which?
3) He gets 9, then we get 6, he gets 7, we get 8, he gets 14, we get 16. Drawish.
6) He gets 8, we pick 7, he gets 9, we pick 3, he gets 16, we pick 14, he gets 12. Maybe winning but I haven't checked if this is the optimal play.
8) Second player gets 9, we get 6, he gets 14, we get 16, he picks 7. Drawish.
9A) Second player gets 7, we get 8, he gets 14, we get 16 and he loses, he cannot defend against 3,8,9 and 2,9,12 while losing all his chances.
9B) Second player gets 8, we get 7, he gets 16, we get 14, he gets 12. This seems promising.
Now, number 210 is the threat. We get 15 and complete it. He lacks 7 and cannot get it. He can't get to 225 either, and there are no number closer or equal distance to 220.
Finally, analyzing what would happen if we start with the other candidate - number 4. Obviously, if he picks anything but 2 we refer to 2,4 case above. So, the options left are similar to the case above, only with players inverted.
If we pick anything but 14 or 16, he gets 14 or 16, depending what we got. We cannot block both (14,16) and (2,8,14)/(2,7,16). 16 fails to picking 8 and 9 (we need to block with 14 and 12, but 3,8,9 remains). 14 is plausible, he cannot make a triplet, we block him. (he 9, we 12), (he 16, we 7), (he 8, we 3)... we end up with 7 again and can make 210 as 3,7,10, winning the game again.
So, the game will end with first player winning with score 210 if both play optimally. There are two possible first moves, after which most other moves are forced.