Basics

By pencil and papers:

$(999999999999)^2=(100000000000-1)^2=1000000000000^2 - 2(1000000000000)(1)+1^2=10000000\ldots(24 \text{zeroes})-2000000000000+1=999999999998000000000000+1= \boxed{999999999998000000000001}$

Therefore, it has 24 decimal digits.

the answer is $24$ since there are $12$ in $999999999999$ and squaring the number means to make the digits of the number twice the multiplicand

We observe:

$9^2=81$ ( $1\implies 2$ )

$99^2=9801$ ( $2\implies 4$ )

So, ( $n\implies 2n$ )

$(999999999999)^2 \implies n=12$

$2n=\boxed{24}$

Ans:

Number of digits
$= \log_{10}(9999999999999)^2 +1 = 2 \times \log_{10}(9999999999999) +1 = 24$