Solution:

The $5$ cells that make up the top row and left column are all leading digits of the three-digit numbers. Therefore, the largest number has leading digit at least $5$, achievable only if $6,7,8$, and $9$ are placed in the bottom right $2 \times 2$ square. Then, the only three-digit numbers with tens digit less than $6$ are the top row and the left column, so unless $5$ is in the top left corner, the three-digit number starting with $5$ will be at least $560$.

Now observe $5$ is next to two other digits; if they are not $1$ or $2$ in some order, then either the top row or left column will read at least $530$. Thus we can assume $5$ is next to $1$ or $2$. The next-smallest remaining digit is $3$, so the three-digit number starting with $52$ must be at least $523$. This is achievable as shown below.