If you want downloads number for each customer, use:
select ssn
, sum(time)
from downloads
group by ssn
If you want just one record -- for a customer with highest number of downloads -- use:
select *
from (
select ssn
, sum(time)
from downloads
group by ssn
order by sum(time) desc
)
where rownum = 1
However if you want to see all customers with the same number of downloads, which share the highest position, use:
select *
from (
select ssn
, sum(time)
, dense_rank() over (order by sum(time) desc) r
from downloads
group by ssn
)
where r = 1