Let's find an integer whose square root has the digit sequence 414213462 immediately following the decimal point.
Let the fractional part given in the problem be denoted by $T$. Let the integer part of the square root of a number that meets the conditions of the problem be denoted by $A$.

The fractional part of the square root of the number we are looking for must be less than

$$
0.414213463 = T + 10^{-9}
$$

Using our notation, the problem can thus be stated as: find an integer $B$ such that

$$
A + T \leq \sqrt{B} \\
10^{-9} \cdot (2A + 2T + 10^{-9}) > 1
$$

which is

$$
2 \cdot 10^{-9} \cdot A + 2 \cdot 10^{-9} \cdot T + 10^{-18} > 1
$$

Choose $A$ such that $2 \cdot 10^{-9} \cdot A$ is exactly 1: $A = 5 \cdot 10^{8}$. (Of course, any integer larger than this is also suitable.)

$A + T = 5 \cdot 10^{8} + 0.414213462$. The integer that is just larger than $(A + T)^2$ is certainly a solution: $\left[(A + T)^2\right] + 1$ (the ceiling of $(A + T)^2$, denoted by $\left\lceil (A + T)^2 \right\rceil$) is a possible solution:

$$
B = 25 \cdot 10^{16} + 2 \cdot 5 \cdot 10^{8} \cdot 414213462 \cdot 10^{-9} + 1 = 25 \cdot 10^{16} + 414213463
$$

Nowhere did we use any special property of $T$, so the solution can be generalized to any fractional part consisting of $k$ digits:

$$
\begin{gathered}
B = 25 \cdot 10^{2k-2} + T \cdot 10^{k} + 1 \\
5 \cdot 10^{k-1} + T \leq \sqrt{B} < 5 \cdot 10^{k-1} + T + 10^{-k}
\end{gathered}
$$