They ask for a two-digit number such that when it is written three times next to itself and then a 1 is added next to it, we get a 7-digit number which is a perfect cube.
Let $x$ be the sought number and $n^{3}$ the perfect cube consisting of 7 digits. Then

$$
101010 x+1=n^{3}
$$

and

$$
10^{6} \leq n^{3}<10^{7}
$$

From 2), it follows that $n$ lies between 100 and $100 \sqrt[3]{10}$, or between 100 and 215.

From 1), it follows that $n^{3}$ ends in 1, which is only possible if $n$ also ends in 1. Furthermore, the equation under 1) can be rewritten as:

$$
101010 x=n^{3}-1
$$

or

$$
101010 x=(n-1) n(n+1)+n-1
$$

The left side of 3) and the first term on the right side are divisible by 30. Indeed, $101010=30 \cdot 3367$, $n-1$ ends in 0 because $n$ ends in 1, and among three consecutive numbers, one is always divisible by 3.

Therefore, $n-1$ must also be divisible by 30. That is,

$$
n-1=30 k
$$

The multiples of 30 that lie between 99 and 214 are:

$$
120, \quad 150, \quad 180, \quad 210
$$

Thus, $n$ must be sought among the following:

$$
\text { 121, } \quad 151, \quad 181, \quad 211
$$

The cubes of these numbers are:

1771561,

3442951,

5929741,

9393931.

We see that only $n=211$ satisfies the problem, and the sought $x=93$.