We note that:
$\dfrac1{1 \times 3 \times 5} = \dfrac1{4} \cdot \left(\dfrac{5-1}{1 \times 3 \times 5} \right)$
$\qquad\qquad\;\; = \dfrac{1}{4} \cdot\left ( \dfrac1{1 \times 3} - \dfrac1{3 \times 5} \right)$
Now, we can rewrite the original series as a Telescoping series and simplify as follows:
$\require{cancel}
\begin{align}
& \left ( \color{blue}{\frac1{1 \times 3 \times 5}} + \color{red}{\frac1{3 \times 5 \times 7}} + \color{#0af}{\frac1{5 \times 7 \times 9}} + \cdots \right ) \\[1em]
=& \small \frac 1 4 \cdot \left ( \color{blue}{\frac1{1 \times 3} - \frac1{3 \times 5}} + \color{red}{\frac1{3 \times 5} - \frac1{5 \times 7}} +\color{#0af}{\frac1{5 \times 7} - \frac1{7 \times 9}} + \cdots \right )\\[1em]
=& \small \frac 1 4 \cdot \left ( \color{blue}{\frac1{1 \times 3} - \cancel{\frac1{3 \times 5}}} + \color{red}{\cancel{\frac1{3 \times 5}} - \bcancel{\frac1{5 \times 7}}} +\color{#0af}{\bcancel{\frac1{5 \times 7}} - \cancel{\frac1{7 \times 9}}} + \cdots \right )\\[1em]
=& \frac 1 4 \cdot \left ( \frac1{1 \times 3}\right )\\[1em]
=& \frac1{12}
\end{align}$
Hence $(D)$ is the Answer.