I don't believe that that is correct. The non-clustered index can be larger (require more pages of storage) than the clustered index.

In a case like the index here, the leaf level of the non-clustered index is the same size as the leaf level of the non-clustered index. If there are enough rows the non-leaf levels of the non-clustered index will be bigger than the non-leaf levels of the clustered index simply because the cluster key is smaller than the other index's key, so that the total size of the non-clustered index will be greater than the total size of the clustered index.