Write SQL query to solve given problem: List the student's email with grade of B in a course with difficulty greater than the 80% of average difficulty of all courses.. Keep the solution inside sql tag ```sql [SQL-Query] ```
SELECT T2.email FROM registration AS T1 INNER JOIN student AS T2 ON T1.student_id = T2.student_id INNER JOIN course AS T3 ON T1.course_id = T3.course_id WHERE T1.grade = 'B' GROUP BY T3.diff HAVING T3.diff > AVG(T3.diff) * 0.8