Answer / anish tuladhar

select
distinct salary
from
(
select
DENSE_RANK() over(order by salary desc) as rnk,
salary
from
employee
) a
where
rnk = 4

Answer / prabhjeet singh sethi

select * from
(select rank(salary) over (partition by employee order by salary desc) as top_salary, employee from table
group by employee)
where top_salary = 4

Answer / icedrop

 select top 1 salary from (select distinct top 4 Salary from tablename order by salary desc ) result order by salary 

Answer / santhosh

;with result as
(
--use dense_rank instead of row_number or rank
select salary, DENSE_RANK() over (order by salary desc) as denserank
from Employees
)
select top 1 * from result
where denserank = 4

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