Solution:

Option (C) $-3 / \sqrt{10}$ is the correct.

Explanation:

As per the question,

It is given that, $\tan \theta=3$ and $\theta$ lies in the third quadrant

$\Rightarrow \cot \theta=1 / 3$

It is known that,

$\begin{array}{l}
\operatorname{Cosec}^{2} \theta=1+\cot ^{2} \theta \\
=1+\left(\frac{1}{3}\right)^{2}=1+\frac{1}{9}=\frac{10}{9} \\
\Rightarrow \sin ^{2} \theta=\frac{9}{10} \\
\Rightarrow \sin \theta=\pm \frac{3}{\sqrt{10}} \\
\Rightarrow \sin \theta=-\frac{3}{\sqrt{10}}, \text { as } \theta \text { lies in third quadrant. }
\end{array}$

As a result, option (C) $-3 / \sqrt{10}$ is the correct answer.