Oracle 21c

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--Source: Markus Winand of modern-sql.com/
--For each country, what city has the highest population? (where the country has more than one city)
--Include the city name as a column.

-- A so-called "top-n per group" problem. Two common solutions
-- 1) ROW_NUMBER() OVER(PARTITON BY)
-- 2) LATERAL (ORDER BY ... FETCH FIRST 1 ROW ONLY)
--    lateral approach needs a countries table, thus skipping here

-- NOTE: your example data is non-deterministic regarding "biggest" (there is a tie)
--       Thus a difference a2 vs a3 or b2 vs b3 is not an error.
--       With this approach it would be even easy to list both cities if there is a tie:
--       Just replace ROW_NUMBER by RANK and it will list the highest pop city
--       along with all other cities that have the same population (ties).

with cities (city, country, population) as (
select 'a1', 'a', 1 from dual union all
select 'a2', 'a', 3 from dual union all
select 'a3', 'a', 3 from dual union all
select 'b1', 'b', 1 from dual union all
select 'b3', 'b', 3 from dual union all
select 'b2', 'b', 3 from dual union all
select 'c1', 'c', 1 from dual
)
SELECT *
  FROM (SELECT cities.*
             , ROW_NUMBER() OVER(PARTITION BY country ORDER BY population DESC) rn
             , COUNT(*)     OVER(PARTITION BY country) ct_cities
          FROM cities
       ) t
 WHERE t.rn = 1
   AND t.ct_cities > 1

CITY	COUNTRY	POPULATION	RN	CT_CITIES
a2	a	3	1	3
b2	b	3	1	3