1. For a body rolling without slipping,

$v = \omega R$

So the two parts of kinetic energy are:

• Translational kinetic energy: $K_t = \frac{1}{2}mv^2$

• Rotational kinetic energy: $K_r = \frac{1}{2}I\omega^2$

Using $\omega = \dfrac{v}{R}$,

$K_r = \frac{1}{2}I\left(\frac{v}{R}\right)^2 = \frac{1}{2}\frac{I}{R^2}v^2$

2. Given that rotational kinetic energy is $50\%$ of translational kinetic energy:

$K_r = 0.5\,K_t$

So,

$\frac{K_r}{K_t} = \frac{1}{2}$

Now,

$\frac{K_r}{K_t} = \frac{\frac{1}{2}\frac{I}{R^2}v^2}{\frac{1}{2}mv^2} = \frac{I}{mR^2}$

Hence,

$\frac{I}{mR^2} = \frac{1}{2}$

3. Compare with standard moments of inertia:

• Solid sphere: $I = \frac{2}{5}mR^2 \quad \Rightarrow \quad \frac{I}{mR^2} = \frac{2}{5}$

• Solid cylinder: $I = \frac{1}{2}mR^2 \quad \Rightarrow \quad \frac{I}{mR^2} = \frac{1}{2}$

• Hollow cylinder: $I = mR^2 \quad \Rightarrow \quad \frac{I}{mR^2} = 1$

• Ring: $I = mR^2 \quad \Rightarrow \quad \frac{I}{mR^2} = 1$

4. The required value is

$\frac{I}{mR^2} = \frac{1}{2}$

This matches a solid cylinder.

Therefore, the correct option is:

$\boxed{\text{B: Solid cylinder}}$