Let $a=7cm$
$b=4\sqrt 3$
$c=\sqrt{13}$cm
$\Rightarrow$ Smallest side is C
Smallest angle will be C
$\cos C=\large\frac{a^2+b^2-c^2}{2ab}$
$\qquad=\large\frac{(7)^2+(4\sqrt 3)^2-(\sqrt{13})^2}{2\times 7\times 4\sqrt 3}$
$\qquad=\large\frac{49+48-13}{56\sqrt 3}$
$\qquad=\large\frac{97-13}{56\sqrt 3}$
$\qquad=\large\frac{\sqrt 3}{2}$
$C=\large\frac{\pi}{6}$
Hence (a) is the correct option.