Answer
$$0.366 \mathrm{s}
$$
Work Step by Step
We use Eq. $15-29$ and the parallel-axis theorem $$I=I_{\mathrm{cm}}+m h^{2}$$ where $h=d .$ For a solid disk of mass $m,$
We know that the rotational inertia about its center of mass will be $$I_{\mathrm{cm}}=m R^{2} / 2 .$$ Therefore,
$$
T=2 \pi \sqrt{\frac{m R^{2} / 2+m d^{2}}{m g d}}=2 \pi \sqrt{\frac{R^{2}+2 d^{2}}{2 g d}}=2 \pi \sqrt{\frac{(2.35 \mathrm{cm})^{2}+2(1.75 \mathrm{cm})^{2}}{2\left(980 \mathrm{cm} / \mathrm{s}^{2}\right)(1.75 \mathrm{cm})}}=0.366 \mathrm{s}
$$
