Answer
The speed of the box is
$$v_{\text{box}}=4.76 \text{ m/s}$$
Work Step by Step
From the law of conservation of energy it follows that Loss of $U$ of box equals gain in $K$ of system. Both the cylinder and
pulley have kinetic energy of the form $K=\dfrac{1}{2}I\omega^2$.
So we can write down:
$m_{\text{box}}gh=\dfrac{1}{2}m_{\text{box}}v^2_{\text{box}}+\dfrac{1}{2}I_{\text{pulley}}\omega^2_{\text{pulley}}+\dfrac{1}{2}I_{\text{cylinder}}\omega^2_{\text{cylinder}}$
and
$\omega_{\text{pulley}}=\dfrac{v_{\text{box}}}{r_{\text{pulley}}},$
$\omega_{\text{cylinder}}=\dfrac{v_{\text{box}}}{r_{\text{cylinder}}}.$
So we have
$m_{\text{box}}gh=\dfrac{1}{2}m_{\text{box}}v^2_{\text{box}}+\dfrac{1}{2}\left(\dfrac{1}{2}m_{\text{pulley}}r^2_{\text{pulley}}\right)\left(\dfrac{v_{\text{box}}}{r_{\text{pulley}}}\right)^2+\dfrac{1}{2}\left(\dfrac{1}{2}m_{\text{cylinder}}r^2_{\text{cylinder}}\right)\left(\dfrac{v_{\text{box}}}{r_{\text{cylinder}}}\right)^2$
and
$v_{\text{box}}=\left(\dfrac{m_{\text{box}}gh}{\frac{1}{2}m_{\text{box}}+\frac{1}{4}m_{\text{pulley}}+\frac{1}{4}m_{\text{cylinder}}}\right)^{1/2}$
$v_{\text{box}}=\left(\dfrac{3\cdot9.8\cdot2.5}{1.50+\frac{1}{4}\cdot7.0}\right)=4.76 \text{ m/s}$
