select university,difficult_level,count(difficult_level)/count(distinct a.device_id) avg_answer_cnt
from user_profile a
inner join (
    select b.device_id,c.difficult_level
from question_practice_detail b
inner join question_detail c
on b.question_id = c.question_id
) aa
on a.device_id = aa.device_id
group by university,difficult_level