Solution:
Denote this number by $N$. Then $N$ can not contain the digits $0, 5, 7$ and its digits must be written in increasing order. Suppose that $N$ has $x_{1}$ ones, $x_{2}$ twos, $x_{3}$ threes, $x_{4}$ fours, $x_{6}$ sixes, $x_{8}$ eights and $x_{9}$ nines, then
$$x_{1} + 2x_{2} + 3x_{3} + 4x_{4} + 6x_{6} + 8x_{8} + 9x_{9} = 2011.$$
(1)

The product of digits of the number $N$ is a power of $6$ when we have the relation
$$x_{2} + 2x_{4} + x_{6} + 3x_{8} = x_{3} + x_{6} + 2x_{9},$$
hence
$$x_{2} - x_{3} + 2x_{4} + 3x_{8} - 2x_{9} = 0.$$
Denote by $S$ the number of digits of $N$ ($S = x_{1} + x_{2} + \ldots + x_{9}$). In order to make the coefficients of $x_{8}$ and $x_{9}$ equal, we multiply relation (1) by $5$, then we add relation (2). We get
$$43x_{9} + 43x_{8} + 30x_{6} + 22x_{4} + 14x_{3} + 11x_{2} + 5x_{1} = 10055$$
which gives
$$43S = 10055 + 13x_{6} + 21x_{4} + 29x_{3} + 32x_{2} + 38x_{1}.$$
Then $10055 + 13x_{6} + 21x_{4} + 29x_{3} + 32x_{2} + 38x_{1}$ is a multiple of $43$ not less than $10055$. The least such number is $10062$, but the relation $10062 = 10055 + 13x_{6} + 21x_{4} + 29x_{3} + 32x_{2} + 38x_{1}$ means that among $x_{1}, x_{2}, \ldots, x_{6}$ there is at least one positive, so
$$10062 = 10055 + 13x_{6} + 21x_{4} + 29x_{3} + 32x_{2} + 38x_{1} \geq 10055 + 13 = 10068$$
which is obviously false. The next multiple of $43$ is $10105$ and from relation
$$10105 = 10055 + 13x_{6} + 21x_{4} + 29x_{3} + 32x_{2} + 38x_{1}$$
we get
$$50 = 13x_{6} + 21x_{4} + 29x_{3} + 32x_{2} + 38x_{1}.$$
By writing as
$$13x_{6} + 21(x_{4} - 1) + 29(x_{3} - 1) + 32x_{2} + 38x_{1} = 0,$$
we can easily see that the only possibility is $x_{1} = x_{2} = x_{6} = 0$ and $x_{3} = x_{4} = 1$. Then $S = 235$, $x_{8} = 93$, $x_{9} = 140$. Since $S$ is strictly minimal, we conclude that
$$N = 34\underbrace{88 \ldots 8}_{93}\underbrace{99 \ldots 9}_{140}.$$