2020 AMC 12A Problem 8

The median is the average of the $2020$th and $2021$st smallest values. The perfect squares that are at most $2020$ are $1^2, \ldots, 44^2$ (since $44^2 = 1936$ and $45^2 = 2025$), so there are $44$ of them. Among the list, the numbers $\le 1976$ are the $1976$ integers $1, \ldots, 1976$ together with those $44$ squares, totaling $1976 + 44 = 2020.$ Thus the $2020$th value is $1976$ and the $2021$st value is $1977,$ making the median $\dfrac{1976 + 1977}{2} = 1976.5.$ Thus, C is the correct answer.