oracle

select max(salary*months), count(*) from employee where salary*months=(select max(salary*months) as s from employee) ;

select max(months*salary) as total_earning, count(*) from Employee where months*salary=(select max(months*salary) from Employee )

Use MySQL:

SELECT (months * salary) AS earnings, COUNT(*)
FROM employee
GROUP BY earnings
ORDER BY earnings DESC
LIMIT 1; 

SELECT 
    CONCAT(max_val.max_earnings, ' ', COUNT(*)) AS result
FROM employee e
JOIN (
    SELECT MAX(months * salary) AS max_earnings
    FROM employee
) AS max_val ON e.months * e.salary = max_val.max_earnings
GROUP BY max_val.max_earnings;

`SELECT months * salary AS earnings, COUNT(*) FROM Employee GROUP BY earnings ORDER BY earnings DESC LIMIT 1;