Solution:

Given equations are
$x^{2}+y^{2}=16 \dots \dots (1)$
$y^{2}=6x \dots \dots (2)$
The area bounded by the circle and parabola
$=2[\text { Area }(\mathrm{OADO})+\text { Area }(\mathrm{ADBA})]$
$=2[$ Area $(\mathrm{OADO})+$ Area $(\mathrm{ADBA})]$
$=2\left[\int_{0}^{2} \sqrt{16 x} d x+\int_{2}^{4} \sqrt{16-x^{2}} d x\right]$
$=2\left[\sqrt{6}\left\{\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right\}_{0}^{2}\right]+2\left[\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_{2}^{4}$
$=2 \sqrt{6} \times \frac{2}{3}\left[x^{\frac{3}{2}}\right]_{0}^{2}+2\left[8 \cdot \frac{\pi}{2}-\sqrt{16-4}-8 \sin ^{-1}\left(\frac{1}{2}\right)\right]$
$=\frac{4 \sqrt{6}}{3}(2 \sqrt{2})+2\left[4 \pi-\sqrt{12}-8 \frac{\pi}{6}\right]$
$=\frac{16 \sqrt{3}}{3}+8 \pi-4 \sqrt{3}-\frac{8}{3} \pi$
$=\frac{4}{3}[4 \sqrt{3}+6 \pi-3 \sqrt{3}-2 \pi]$
$=\frac{4}{3}[\sqrt{3}+4 \pi]$
$=\frac{4}{3}[4 \pi+\sqrt{3}]$ units
Area of circle $=\pi(r)^{2}$
$=\pi(4)^{2}$
$=16 \mathrm{n}$ units
$\therefore$ Required area $=16 \pi-\frac{4}{3}[4 \pi+\sqrt{3}]$
$=\frac{4}{3}[4 \times 3 \pi-4 \pi-\sqrt{3}]$
$=\frac{4}{3}(8 \pi-\sqrt{3})$ sq.units
As a result, the correct answer is $\mathrm{C}$.