b
$T=2 \pi \sqrt{\frac{I_{\text {support}}}{m g l_{\text {cm}}}}$
$I_{\text {support}}=m\left(l \sin 60^{\circ}\right)^{2}=\frac{3}{4} m l^{2}$
$l_{c m}=l \sin 60^{\circ}=\frac{\sqrt{3} l}{2}$
$\Rightarrow T=2 \pi \sqrt{\frac{\left(\frac{3}{4} m l^{2}\right)}{m g\left(\frac{\sqrt{3} l}{2}\right)}}=2 \pi \sqrt{\frac{\sqrt{3} L}{2 g}}$