Relationship between an integer N and the number of bits n required to represent the integer
The integer $N$ takes $\log_2 N$ bits to write down.  Set $n = \log_2 N$.  Then $N = 2^n$, so $\sqrt{N} = 2^{n/2}$.

I'd guess they are using the natural logarithm (to base $e$), which is where they get $e$ instead of $2$.  That's a bit odd but I suppose it is valid.

See also Complexity of multiplication and Precise runtime of the algorithm to find number of digits in an integer.