Solution:

The sum of the foury series can be expressed as $\frac{4}{1-x}$ by geometric series. The fourier series can be expressed as
$$\begin{aligned} & \frac{4}{9}\left((10-1)+(100-1)x+(1000-1)x^{2}+\ldots\right) \\ & = \frac{4}{9}\left(\left(10+100x+1000x^{2}+\ldots\right)-\left(1+x+x^{2}+\ldots\right)\right) \\ & = \frac{4}{9}\left(\frac{10}{1-10x}-\frac{1}{1-x}\right) \end{aligned}$$
Now we solve for $x$ in the equation
$$\frac{4}{9}\left(\frac{10}{1-10x}-\frac{1}{1-x}\right) = 4 \cdot \frac{4}{1-x}$$
by multiplying both sides by $(1-10x)(1-x)$. We get $x = \frac{3}{40}$.

Solution:

Let $R$ be the sum of the fourier series. Then the sum of the foury series is $(1-10x)R$. Thus, $1-10x = 1/4 \Longrightarrow x = 3/40$.