Answer
The period of the physical pendulum is $0.54s$
Work Step by Step
1) When the pendulum is a physical pendulum, we have $$2\pi f=\frac{2\pi}{T}=\sqrt{\frac{mgL}{I}}$$ $$T=\frac{2\pi}{\sqrt{\frac{mgL}{I}}}=\sqrt{\frac{4\pi^2I}{mgL}}$$
The physical pendulum is a thin rod, whose rotation axis is at one end, so $I_1=\frac{1}{3}mL^2$. The rod's center of gravity is at its center, since the rod is uniform. Therefore, if the rod's length is $d$, then $L=d/2$. $$T_1=\sqrt{\frac{4\pi^2mL^2}{3mgL}}=\sqrt{\frac{4\pi^2L}{3g}}$$ $$T_1=\sqrt{\frac{4\pi^2(d/2)}{3g}}=\sqrt{\frac{2\pi^2d}{3g}}$$
2) When the pendulum is a simple pendulum (the sphere is still attached), we have $$2\pi f = \frac{2\pi}{T}=\sqrt{\frac{g}{L}}$$ $$T=\sqrt{\frac{4\pi^2L}{g}}$$
The pendulum's length now is the rod's length $L=d$. $$T_2=\sqrt{\frac{4\pi^2d}{g}}$$
Therefore, $$\frac{T_1}{T_2}=\sqrt{\frac{\frac{2\pi^2L}{3g}}{\frac{4\pi^2L}{g}}}=\sqrt{\frac{\frac{2}{3}}{1}}=\sqrt{\frac{2}{3}}=0.82$$ The period when the sphere is still on $T_2=0.66s$, so $$T_1=0.54s$$
