The correct answer is $\boxed{\frac{{2\pi }}{{\sqrt {10} }}}$.
To solve this integral, we can use the substitution $u=\sin\theta$. This gives us $\text{d}u=\cos\theta\,\text{d}\theta$, so we can rewrite the integral as
$$\int_0^{2\pi } {\left( {\frac{3}{{9 + {{\sin }^2}\theta }}} \right){\text{d}}\theta } = \int_0^1 {\left( {\frac{3}{{9 + u^2 }}} \right){\cos \theta\,\text{d}}\theta }$$
We can now evaluate this integral using the following trigonometric identity:
$$\int_0^1 {\frac{du}{{9 + u^2 }}} = \frac{\pi}{\sqrt{10}}$$
Substituting this back in, we get
$$\int_0^{2\pi } {\left( {\frac{3}{{9 + {{\sin }^2}\theta }}} \right){\text{d}}\theta } = \frac{3\pi}{\sqrt{10}} = \frac{{2\pi }}{{\sqrt {10} }}$$
Therefore, the value of the integral is $\boxed{\frac{{2\pi }}{{\sqrt {10} }}}$.
Here is a brief explanation of each option:
- Option A: $\frac{{2\pi }}{{\sqrt {10} }}$. This is the correct answer.
- Option B: $2\sqrt {10} \pi$. This is incorrect because it is not equal to $\frac{{2\pi }}{{\sqrt {10} }}$.
- Option C: $\sqrt {10} \pi$. This is incorrect because it is not equal to $\frac{{2\pi }}{{\sqrt {10} }}$.
- Option D: $2\pi$. This is incorrect because it is not equal to $\frac{{2\pi }}{{\sqrt {10} }}$.