Answer
The block attached to the solid cylinder reaches the ground first.
Work Step by Step
We know that the torque applied by the tension force in the rope attached to the block is given by
$$\tau= TR\sin90^\circ=TR$$
where when at rest, $T=mg$, where $m$ is the block's mass.
$$\tau= m_{ }gR $$
We also know that the net torque is given by
$$\tau=I\alpha$$
Thus,
$$I\alpha= m_{ }gR $$
Hence, the angular acceleration is given by
$$ \alpha= \dfrac{m_{ }gR }{I}$$
This is a general formula for both cases.
So, in the Solid cylinder case,
$$ \alpha_1= \dfrac{m_{ }gR }{I_1}$$
where $I_1=\frac{1}{2}MR^2$ where $M$ is the cylinder's mass.
$$ \alpha_1= \dfrac{m g\color{red}{\bf\not} R }{\frac{1}{2}MR^{\color{red}{\bf\not} 2}}=\color{blue} {\dfrac{2mg}{MR}}$$
So, in the shell cylinder case,
$$ \alpha_2= \dfrac{m_{ }gR }{I_2}$$
where $I_2= MR^2$ where $M$ is the cylinder's mass.
$$ \alpha_1= \dfrac{m g\color{red}{\bf\not} R }{ MR^{\color{red}{\bf\not} 2}}=\color{blue} {\dfrac{ mg}{MR}}$$
From the two blue results, we can see that
$$\alpha_1=2\alpha_2$$
This means that the angular acceleration of the solid cylinder is greater than the shell cylinder by two times. This means that the solid one rotates faster and hence the block falls faster. Therefore the block attached to the solid cylinder reaches the ground first.
