2026/005) How do I the smallest four-digit integer N, such that $\sqrt{3\sqrt{N}}$ is an integer?

We have one 3 under the square root so we should have another 3 that should be part of $\sqrt{N}$ So it should be $N= (3x)^2$ and x has to be a square so smallest 4 digit number giving $N= 9x^4$ satisfying that it is smallest 4 digit number    when $x=4$ giving  $N= 9 * 256$ or 2304