Answer
$\approx 1.00002$
Work Step by Step
a) We know that the angular frequency for a physical pendulum is given by
$$\omega=2\pi f=\sqrt{\dfrac{Mgl}{I}}$$
where we know that $f=1/T$, so
$$ \dfrac{2\pi }{T}=\sqrt{\dfrac{Mgl}{I}}$$
Hence, the period of a physical pendulum is given by
$$ T =2\pi \sqrt{\dfrac{I}{Mgl}}$$
where in our case, $l=L$, as shown in the figure below.
$$ T =2\pi \sqrt{\dfrac{I}{MgL}}\tag 1$$
So we have to find the moment of inertia of our system and we are going to use the parallel axis theorem.
$$I=I_{\rm cm,sphere}+Md^2$$
where $D=L$, and $I_{\rm cm,sphere}=\frac{2}{5}MR^2$
$$I=\frac{2}{5}MR^2+ML^2$$
Plugging into (1);
$$ T =2\pi \sqrt{\dfrac{\frac{2}{5}\color{red}{\bf\not} MR^2+\color{red}{\bf\not} ML^2}{\color{red}{\bf\not} MgL}} $$
$$ \boxed{T =2\pi \sqrt{\dfrac{\frac{2}{5} R^2+ L^2}{ gL}}}$$
---
b) We need to find the ratio of $T_{\rm real}/T_{\rm simple}$
where the real one is on the boxed equation above and the simple is $T=2\pi \sqrt{L/g}$
$$\dfrac{T_{\rm real}}{T_{\rm simple}}=\dfrac{\color{red}{\bf\not} 2\color{red}{\bf\not} \pi \sqrt{\dfrac{\frac{2}{5} R^2+ L^2}{ gL}}}{\color{red}{\bf\not} 2\color{red}{\bf\not} \pi \sqrt{\dfrac{L}{g}}}$$
$$\dfrac{T_{\rm real}}{T_{\rm simple}}= \sqrt{\dfrac{\frac{2}{5} R^2+ L^2}{ \color{red}{\bf\not} gL}\times \dfrac{\color{red}{\bf\not} g}{L}}$$
$$\dfrac{T_{\rm real}}{T_{\rm simple}}= \sqrt{\dfrac{\frac{2}{5} R^2+ L^2}{ L^2} }$$
$$\dfrac{T_{\rm real}}{T_{\rm simple}}= \dfrac{\sqrt{\frac{2}{5} R^2+ L^2}}{ L } $$
Plugging the known;
$$\dfrac{T_{\rm real}}{T_{\rm simple}}= \dfrac{\sqrt{\frac{2}{5} (1\times 10^{-2})^2+ (1)^2}}{ 1 } =\color{red}{\bf 1.0002} $$
This means that the simple pendulum approximation is fair enough in many cases.
