, 2023

Since $2023 = 7 \times 17^2$,
then any perfect square that is a multiple of 2023 must have prime
factors of both 7 and 17.

Furthermore, the exponents of the prime factors of a perfect square must
be all even.

Therefore, any perfect square that is a multiple of 2023 must be
divisible by $7^2$ and by $17^2$, and so it is at least $7^2 \times 17^2$ which equals $7 \times 2023$.

Therefore, the smallest perfect square that is a multiple of 2023 is
$7 \times 2023$.

We can check that $2023^2$ is larger
than $7 \times 2023$ and that none of
$4 \times 2023$ and $17 \times 2023$ and $7 \times 17 \times 2023$ is a perfect
square.