5. (8 points) Rearrange the four digits of 2016 to form a four-digit perfect square; what is this four-digit perfect square? $\qquad$ .
【Solution】Solution: According to the analysis, rearrange the four digits of 2016, let this four-digit number be $A=n^{2}$,
$$
32^{2}<1026 \leqslant A \leqslant 6210<80^{2}, 32<n<80,
$$

To form a four-digit perfect square, the unit digit must be $0,1,6$,
Since when the unit digit is 0, the four-digit number must have two 0s to be a square number, thus numbers with a unit digit of 0 and 2 can be excluded,

Numbers with a unit digit of 1 and 6 are: $2061, 2601, 6021, 6201, 1206, 1026, 2016, 2106$, a total of eight numbers,

Among them, if the unit digit is 6, then $n=36, 46, 56, 66, 76$, and $36^{2}=1296, 46^{2}=2116, 56^{2}=3136$, $66^{2}=4356, 76^{2}=5776$, none of which meet the criteria, so they are excluded,

Therefore, the unit digit is 1, and among $2061, 2601, 6021, 6201$, only $2601=51^{2}$, is a square number, this four-digit number is 2601,

Therefore, the answer is: 2601.