My simple solution using SQL Server:

SELECT S.Name
FROM Students S WITH (NOLOCK)
INNER JOIN Friends F WITH (NOLOCK) ON S.ID = F.ID
INNER JOIN Packages P WITH (NOLOCK) ON P.ID = S.ID
INNER JOIN Packages PF WITH (NOLOCK) ON PF.ID = F.Friend_ID AND P.Salary < PF.Salary
ORDER BY PF.Salary

MS SQL Server

WITH CTE AS( select s.id,s.name,f.friend_id,p.salary as salary, q.salary as friendssalary from students s INNER JOIN Friends f ON s.id=f.id INNER JOIN packages p ON s.id=p.id INNER JOIN packages q ON f.friend_id=q.id
) select name from CTE where friendssalary>salary order by friendssalary

In MySql

(select A.name from
    (select a.id id, a.name name, p.salary salary, f.friend_id fid from Students a 
    join friends f on f.id=a.id
    join packages p on p.id=a.id) A
join students b on b.id=A.fid
join packages pp on pp.id=A.fid
where A.salary<pp.salary
order by pp.salary)

In MS SQL

with t as (select stu.id as ID, stu.name as Name , pak.salary as Salary , frnd.friend_id as FriendId from students as stu , friends as frnd , packages as pak where stu.id = pak.id and frnd.id = pak.id ),

s as ( select t1.ID as tid , t1.Name as Names , t1.Salary as firstsal , t1.FriendId , t2.Salary as secondsal from t as t1 , t as t2 where t1.FriendId = t2.ID )

select Names from s where firstsal < secondsal order by secondsal;

simple in mysql, use alias:

select s.name from students s inner join friends f on s.id=f.id inner join packages p on p.id=s.id inner join packages p1 on p1.id=f.friend_id where (p1.salary-p.salary)>0 order by p1.salary;

Oracle solution:

select s.name from students s, friends f, packages p, packages p2 where s.id = f.id and f.friend_id = p2.id and s.id = p.id and p.salary < p2.salary order by p2.salary;

Maybe I did not understand the following line of the problem statement

Friends contains two columns: ID and Friend_ID (ID of the ONLY best friend). ... but the correct solution treats Friends relation as unidirectional while it is bidirectional.

Think of Amanda #16 who earns 21.1 which is less than #19 who earns 33.330002

Same with Dyana #5 earning 31.5 which is less than #6 earning 45.0

Same with Jenny #6 earnning 45 which is less than #7 who earns 47

There are some students like Scarlet #14 who has two friends with better compensation and it's not clear which of two salaries to order by:

"Scarlet #14 earning 15.1 which is less than Priya #9 earning 39" is a part of the correct answer but" Scarlet #14 earning 15.1 which is less than ... #1 earning 15.5" does not belong to a correct answer.

Nice Oracle solution:

SELECT  std.name
FROM    students std
,             (   SELECT  frd.id, (SELECT salary FROM packages WHERE id = frd.id) AS my_salary
                        ,       frd.friend_id, (SELECT salary FROM packages WHERE id = frd.friend_id) AS friend_salary
                        FROM    friends frd) X
WHERE   X.friend_salary > X.my_salary
AND       X.id = std.id
ORDER BY X.friend_salary;

MySQL Solution:

select s.name from students s inner join friends f on s.id = f.id inner join packages p on f.friend_id = p.id where p.salary > (select salary from packages p1 where p1.id = s.id) order by p.salary