MySQL 8.0

By using db<>fiddle, you agree to license everything you submit by Creative Commons CC0.

create table data (
  cust varchar(10), 
  amt integer, 
  rnk integer
  );
insert into data values 
    ('a', 10, 1),
    ('a', 5, 2),
    ('a', 15, 3),
    ('b', 20, 1),
    ('b', 30, 2),
    ('c', 3, 1);

Records: 6  Duplicates: 0  Warnings: 0


with max_rank as (
  select cust, max(rnk) as m_rank
  from data
  where rnk in (1,2)
  group by cust
  )
select d.cust, d.amt, d.rnk
from data d
join max_rank r
  on d.cust = r.cust
 and d.rnk = r.m_rank


cust
amt
rnk


a
5
2

b
30
2

c
3
1