Answer
$\frac{\pi}{3}\sqrt {\frac{58D}{g}}$
Work Step by Step
For a physical pendulum $T=2\pi \sqrt {\frac{I}{mgL}}$ , where T - oscillation period, I - Rotational inertia of the system through the pivot, L - distance from the pivot to center of mass of the ball, g - gravitational acceleration ($9.8\space m/s^{2}$)
$T=2\pi \sqrt {\frac{I}{mgL}}-(1)$
According to the parallel axis theorem we can write,
$I=I_{cm}+mL^{2}$ (Please see the attached image)
$I=\frac{2}{3}m(\frac{D}{2})^{2}+m(D+\frac{D}{2})^{2}= \frac{mD^{2}}{6}+\frac{9mD^{2}}{4}$
$I=\frac{2mD^{2}+27mD^{2}}{12}=\frac{29mD^{2}}{12}-(2)$
(2)=>(1)
$T=2\pi\sqrt {\frac{29mD^{2}/12}{mg\times\frac{3D}{2}}}=2\pi \sqrt {\frac{29D}{18g}}=2\pi \sqrt {\frac{58D}{36g}}$
$T=\frac{2\pi}{6}\sqrt {\frac{58D}{g}}=\frac{\pi}{3}\sqrt {\frac{58D}{g}}$
