Hi Keira!
Great question. The explanation doesn't provide very much detail, unfortunately. 😥
Take a look at B and D, first. Focus on the charges. In B, there are 3(3+) on left and 2(2+) on the right. The charges aren't balanced! Similarly, for D, there are 2(2+) on the left and 3(3+) on the right. The charges also aren't balanced! So, B and D cannot be correct!
For A and C, both have 2(3+) or 3(2+) on the left and right, indicating a balnace of charges. So which one is correct?
Take a look at the table. In the first row and second column, the table tells us that a mixture of Al3+(aq) and Cu(s) gives NO CHANGE. This means that Al3+(aq) DON'T REACT. Took a look at reaction A. If Al3+(aq) and Cu(s) don't react, then how can reaction A proceed? 😮
By process of elimination, only C remains. Click and you're done!
You can confirm that it must be true (if you really want to), by seeing that Cu2+(aq) and Al(s) produces a new solid (second row, first column of the table).
Let me know if that helps. 😀