jmc

We consider it by four cases:

$\bullet$ Case 1: There are $3$ perfect squares that only have $1$ digit, $1^2,$ $2^2,$ and $3^2.$

$\bullet$ Case 2: The smallest perfect square that has $2$ digits is $4^2,$ and the largest is $9^2,$ so that's a total of $6$ perfect squares with $2$ digits.

$\bullet$ Case 3: The smallest perfect square with $3$ digits is $10^2,$ and the largest is $31^2,$ yielding a total of $22.$

$\bullet$ Case 4: The smallest perfect square with $4$ digits is $32^2,$ and the last one that is no greater than $2500$ is $50^2,$ giving a total of $19.$

So we have a total of $1\times 3+2\times 6+3\times 22+4\times 19=\boxed{157}$ digits.