Find the smallest natural number $x$, such that in decimal notation, the number $x^{3}$ has the first three digits and the last four digits as 1.
Solving the problem, the last four digits are only related to the last four digits of $x$.

Number $x^{3}$ ends with 1, so the last digit of number $x$ can only be 1. Cubing the numbers 01, 11, 21, ..., 91, we confirm that only $71^{3}=357911$. Therefore, the last two digits of number $x$ are 71. Cubing the numbers 071, 171, 271, ..., 971, we confirm that only the number $471^{3}=104+87111$. Therefore, the last three digits of $x$ are 471. Similarly, cubing the numbers 0471, 1471, 2471, ..., 9471, we find that the last four digits of number $x$ are 8471.

Let $n$ be the number of digits between 111 and 1111 in $x^{3}$. There are three possible scenarios.
$$
\begin{array}{l}
\text { 1) } n=3 k, k \geqslant 0, \text { then } \\
111 \cdot 10^{3 k+4}<x^{3}=111 \frac{\cdots \cdots 1}{3 k \text { digits }} 111 \\
\therefore 112 \cdot 10^{3 k+4} .
\end{array}
$$

Therefore,
$$
\begin{array}{l}
10.353988 \cdot 10^{k+1} \approx \sqrt[3]{1110} \cdot 10^{k}<x \\
<\sqrt[3]{1120} \cdot 10^{k} \approx 10.384988 \cdot 10^{k+1} .
\end{array}
$$

When $k=0$, it is easy to see that the required number $x$ does not exist. If $k \geqslant 1$, then the required number $x$ starts with 103, i.e., $x=103 \cdots 8471$. The smallest such number is 1038471. It is not difficult to verify that this number satisfies the required conditions.
$$
\begin{array}{l}
\text { 2) } n=3 k+1, k \geqslant 0 \text {, then } \\
111 \cdot 10^{3 k+5}<x^{3}=111 \ldots \ldots \ldots \cdots \cdot 1111 \\
3 k+1 \text { digits } \\
\end{array}
$$
$<112 \cdot 10^{3 k+5}$. Therefore, 22.306991 $10^{k+1}$ $\approx \sqrt[3]{11100} \cdot 10^{k+1}<x<\sqrt[3]{11200} \cdot 10^{k+1}$ $\approx 22.373779 \cdot 10^{k+1}$.

When $k=0$, the required number $x$ does not exist. When $k \geqslant 1$, the required number has the form $x=225 \cdots 8471$, and there is at least one digit in the middle position. However, any number of this form is clearly greater than the (seven-digit) number 1038471. Therefore, none of these numbers satisfy the condition.
3) $n=3 k+2, k \geqslant 0$. This case is discussed similarly to case 2). Similarly, we find numbers greater than 1038471. Therefore, the required number is 1038471.