select university,
        difficult_level,
        count(difficult_level)/count(distinct t1.device_id) as avg_answer_cnt
        from question_practice_detail t1

        join user_profile t2 on t1.device_id=t2.device_id

        join question_detail t3 on t1.question_id=t3.question_id

        where university='山东大学'
        group by difficult_level ;

参照上一题，微调count()，group by，以及增加where条件查询即可。