1. We need to evaluate $I=\int_0^{1/2} \frac{x^2}{(1-x^2)^{3/2}}\,dx$ and compare it with $I=\frac{k}{6}.$

2. Rewrite the integrand in a useful form.

Notice that $x^2=1-(1-x^2),$ so $\frac{x^2}{(1-x^2)^{3/2}}=\frac{1}{(1-x^2)^{3/2}}-\frac{1}{\sqrt{1-x^2}}.$

Thus, $I=\int_0^{1/2}\frac{dx}{(1-x^2)^{3/2}}-\int_0^{1/2}\frac{dx}{\sqrt{1-x^2}}.$

3. Evaluate each integral separately.

We use the standard results: $\int \frac{dx}{(1-x^2)^{3/2}}=\frac{x}{\sqrt{1-x^2}}+C,$ because $\frac{d}{dx}\left(\frac{x}{\sqrt{1-x^2}}\right)=\frac{1}{(1-x^2)^{3/2}}.$

Also, $\int \frac{dx}{\sqrt{1-x^2}}=\sin^{-1}x + C.$

Hence, $I=\left[\frac{x}{\sqrt{1-x^2}}-\sin^{-1}x\right]_0^{1/2}.$

4. Substitute the limits.

At $x=\frac12$: $\frac{x}{\sqrt{1-x^2}}=\frac{1/2}{\sqrt{1-1/4}}=\frac{1/2}{\sqrt{3/4}}=\frac{1}{\sqrt3},$ and $\sin^{-1}\left(\frac12\right)=\frac{\pi}{6}.$

At $x=0$: $\frac{x}{\sqrt{1-x^2}}-\sin^{-1}x=0-0=0.$

Therefore, $I=\frac{1}{\sqrt3}-\frac{\pi}{6}.$

Now, $\frac{1}{\sqrt3}=\frac{\sqrt3}{3}=\frac{2\sqrt3}{6},$ so $I=\frac{2\sqrt3}{6}-\frac{\pi}{6}=\frac{2\sqrt3-\pi}{6}.$

Thus, $\frac{k}{6}=\frac{2\sqrt3-\pi}{6} \implies k=2\sqrt3-\pi.$

5. Compare with the options.

So the correct option is: $\boxed{\text{D: }2\sqrt3-\pi}.$