Pulleys pulleys everywhere!

At any moment $t$ , the small mass $m$ is at height $x(t)$ and velocity $v(t)=\frac{dx}{dt}$ and its kinetic energy is $\frac{1}{2}mv^2$ . The center of the reel must have a velocity = $3v(t)$ giving angular velocity of $\omega =\frac{v(t)}{R}$ .

The kinetic energy of the reel is $\frac{1}{2}(4m)(3v)^2 + \frac{3}{2}mR^2\times \omega^2.$

$m.g.x(t)=\frac{1}{2}mv^2+18mv^2 + \frac{3}{2}mv^2 = 20mv^2.$

$x(t)=\frac{20}{g}.v^2=2v(t)^2$

$\frac{dx(t)}{dt}=4v(t).\frac{dv(t)}{dt}$

$v(t)=4v(t).a(t)$

$a(t)=0.25 \text{ m/s}^2= 25 \text{ cm/s}^2.$

Acceleration of the center of the reel is 3 times the acceleration of the block, therefore,

$a_{\text{reel}} = \boxed{\SI{75}{\centi\meter\per\second\squared}}$ .