It is given that $\text{f}(x) = \frac{1}{\sqrt{1 - 4x^2}}$ where $-\frac{1}{2} < x < \frac{1}{2}$.

1. Rewrite the function in index form to prepare for binomial expansion.

$\text{f}(x) = (1 - 4x^2)^{-\frac{1}{2}}$

2. Substitute the appropriate variable and exponent into the standard binomial series expansion up to the third power.

$\text{f}(x) = 1 + \left(-\frac{1}{2}\right)(-4x^2) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}(-4x^2)^2 + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!}(-4x^2)^3 + \dots$

3. Simplify the coefficients to obtain the final polynomial up to the term in $x^6$.

$f(x) = 1 + 2x^2 + \frac{3}{8}(16x^4) + \left(-\frac{15}{48}\right)(-64x^6) + \dots$
$f(x) \approx 1 + 2x^2 + 6x^4 + 20x^6$

1. Determine the Maclaurin expansion of $\text{f}(x)$ from part (a) (required as a prerequisite).

From part (a), the Maclaurin expansion of $\text{f}(x) = (1 - 4x^2)^{-1/2}$ up to $x^6$ is calculated using the standard binomial series:
$\text{f}(x) = 1 + \left(-\frac{1}{2}\right)(-4x^2) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}(-4x^2)^2 + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!}(-4x^2)^3 + \dots$
$\text{f}(x) \approx 1 + 2x^2 + 6x^4 + 20x^6$

2. Differentiate $\sin^{-1} 2x$ to establish its relationship with $\text{f}(x)$.

$\frac{\text{d}}{\text{d}x} (\sin^{-1} 2x) = \frac{1}{\sqrt{1 - (2x)^2}} \cdot \frac{\text{d}}{\text{d}x}(2x) = \frac{2}{\sqrt{1 - 4x^2}} = 2\text{f}(x)$

3. Use the series expansion of $\text{f}(x)$ to express the derivative as a series.

$\frac{\text{d}}{\text{d}x} (\sin^{-1} 2x) \approx 2(1 + 2x^2 + 6x^4 + 20x^6)$
$\frac{\text{d}}{\text{d}x} (\sin^{-1} 2x) \approx 2 + 4x^2 + 12x^4 + 40x^6$

4. Integrate the series term-by-term and evaluate the constant of integration to find the expansion for $\sin^{-1} 2x$.

$\sin^{-1} 2x = \int (2 + 4x^2 + 12x^4 + 40x^6) \, \text{d}x$
$\sin^{-1} 2x = 2x + \frac{4}{3}x^3 + \frac{12}{5}x^5 + \frac{40}{7}x^7 + C$
To find the integration constant $C$, substitute $x = 0$:
$\sin^{-1}(0) = 0 + 0 + 0 + 0 + C \implies C = 0$
Thus, the first four non-zero terms of the Maclaurin series are $2x + \frac{4}{3}x^3 + \frac{12}{5}x^5 + \frac{40}{7}x^7$.