Give it a thought!

In base $7$ , the greatest number of $2$ digits is $(66)_{7}$ , whose square is $(6501)_{7}$ . Least $3$ digit number in base $7$ is $(100)_{7}$ , whose square is $(10000)_{7}$ . Hence, the greatest perfect square of $4$ digits in base $7$ is $\boxed{ (6501)_{7} }$ .

The greatest 4 digit number in base 7 is $6666_{7)} = 2400_{10)}; \sqrt{2400} = 48.98$ so the greatest 4 digit number which is a perfect square in base 7 is $48_{10)}^2 = 2304_{10)} = 6501_{7)}$