Step 1: Use a substitution:
Let $u = 2x$, so $dx = \frac{du}{2}$.
When $x = 0$, $u = 0$. When $x = 2a$, $u = 4a$.
Step 2: Change the integral:
\[ \int_{0}^{2a}\frac{dx}{1+4x^{2}} = \frac{1}{2}\int_{0}^{4a}\frac{du}{1+u^{2}} = \frac{1}{2}\tan^{-1}(4a) \]
Step 3: Use the given value:
Set $\frac{1}{2}\tan^{-1}(4a) = \frac{\pi}{6}$.
Multiply by 2 to get $\tan^{-1}(4a) = \frac{\pi}{3}$.
This angle lies in the principal range, so we can take the tangent safely.
\[ 4a = \sqrt{3} \]
Step 4: Test the options:
Divide by 4 to get $a = \frac{\sqrt{3}}{4}$.
Option (A) is this value.
Putting $a = \frac{\sqrt{3}}{2}$ gives $4a = 2\sqrt{3}$ and the integral exceeds $\frac{\pi}{6}$.
Options (C) and (D) give even larger values.
Final Answer:
$a = \frac{\sqrt{3}}{4}$, option (A).