First number that squared has 2 times as many digits.

For the square of a number $a$ to have twice as many digits as $a$ but still be minimal, $a$ has to be the smallest integer greater than the square root of the power of 10 that has the required number of digits.

Example:

$a$ is a four digit number. Hence, it's square has to be an 8 digit number, so at least $10^7 = 10,000,000$ . To be minimal, $a$ has to be just above this number's square root, $a = \left\lceil \sqrt{10^7} \right\rceil = 3163$ . And, in fact, $3163^2= 10,00,4569 > 10,000,000$ but $3162^2= 9,998,244 < 10,000,000$ .

In general, for an $n$ digit number, $f(n) = a = \left\lceil \sqrt{10^{2n-1}} \right\rceil$ .

$\displaystyle \begin{aligned} & \lim_{n \to \infty} \frac {f(n)} {10^n} \\ = & \lim_{n \to \infty} \frac {\left\lceil \sqrt{10^{2n-1}} \right\rceil} {10^n} \\ = & \lim_{n \to \infty} \frac {\sqrt{10^{2n-1}}} {10^n} \\ = & \lim_{n \to \infty} \frac {10^{n-\tfrac 12}} {10^n} \\ = & \lim_{n \to \infty} \frac {10^n 10^{-\frac 12}} {10^n} \\ = & \lim_{n \to \infty} 10^{-\frac 12} \\ = & 10^{-\frac 12} = \sqrt{\frac 1{10}} \\ \approx & \boxed{0.3162} \end{aligned}$