The moment of inertia of a cylinder about its own axis is equal to its moment of inertia about an axis passing through its centre and normal to its length. The ratio of length to radius is

Moment of inertia about the x-axis,

$I_X+I_Z=I_Y$

Due to symmetry 

$$\begin{gathered} I_X=I_Z \\ 2I_{X_1}=\frac{1}{2}M{R}^2 \\ {I}_{X_2}=\frac{1}{4}M{R}^2 \end{gathered}$$

The moment of inertia about the centre is because the cylinder is thought of as a rod.

$I_{X_2}=\frac{M{l}^2}{12}$

The inertia of the net moment about the x-axis

$I_x=I_{X1}+I_{X2}=\frac{M{l}^2}{12}+\frac{M{R}^2}{4}$

Radius to length ratio,

$$\begin{gathered} I_X=I_Y \\ \Rightarrow \frac{M{l}^2}{12}+\frac{M{R}^2}{4}=\frac{M{R}^2}{2} \\ \Rightarrow \frac{l^2}{12}=\frac{R^2}{2}-\frac{R^2}{4}=\frac{R^2}{4} \\ \Rightarrow {\left(\frac{l}{R}\right)}^2=\frac{12}{4}=\frac{3}{1} \\ \Rightarrow l:R=\sqrt{3}:1 \end{gathered}$$

Hence the correct answer is $\sqrt{3}:1.$