select a.university, c.difficult_level, 
round(count(b.id)/count(distinct(a.device_id)),4) avg_answer_cnt
    from user_profile a
        join question_practice_detail b on a.device_id = b.device_id
        join question_detail c on b.question_id = c.question_id
    group by a.university, c.difficult_level;

需要注意，对a.device_id去重，否则出现平均值都是1的错误。