Answer
$1.11\;\rm s$
Work Step by Step
First of all, we need to draw the force diagram of the three objects, as seen below. To find the time the left block takes to reach the ground, we need to find its acceleration.
We assume that the rope that connects the two blocks is nonstretchable, so the two blocks must have the same acceleration magnitude but one is accelerating upward while the other is accelerating downward.
Now we need to apply Newton's second law on the left block where we chose downward to be the positive direction.
$$\sum F_y=m_1g-T_1=m_1a_y$$
Hence,
$$T_1=m_1g-m_1a_y\tag 1$$
Applying Newton's second law on the right block where we chose upward to be the positive direction.
$$\sum F_y=T_2-m_2g=m_2a_y$$
Hence,
$$T_2=m_2a_y+m_2g\tag 2$$
Applying Newton's second law on the pulley where we chose counterclockwise to be the positive direction.
$$\sum \tau=T_1R-T_2R-fR=I\alpha $$
We are given the torque done by the friction (0.50 N.m) which must oppose the direction of the rotation of the pulley that's why we chose its direction to be affecting clockwise.
Hence,
$$T_1R-T_2R-0.50 =I\alpha $$
We know that the moment of inertia of the pulley, which is as same as that of a disk, is given by $I=\frac{1}{2}MR^2$.
$$T_1R-T_2R-0.50 =\frac{1}{2}MR^2\alpha $$
$$R(T_1-T_2)-0.50 =\frac{1}{2}MR^2\alpha $$
Plugging from (1) and (2);
$$R\left(m_1g-m_1a_y-[m_2a_y+m_2g]\right)-0.50 =\frac{1}{2}MR^2\alpha $$
$$R\left(m_1g-m_1a_y- m_2a_y-m_2g \right)-0.50 =\frac{1}{2}MR^2\alpha $$
Recalling that $\alpha=a_T/R$ where $a_T$ is the tangential acceleration which is here $a_y$. Thus, $\alpha=\dfrac{a_y}{R}$.
Thus,
$$R\left((m_1 -m_2)g-(m_1 + m_2)a_y \right)-0.50 =\frac{1}{2}MR^{\color{red}{\bf\not} 2}\dfrac{a_y}{\color{red}{\bf\not} R} $$
Dividing by $R$;
$$ (m_1 -m_2)g-(m_1 + m_2)a_y-\dfrac{0.50 }{R} =\frac{1}{2}M a_y$$
$$ (m_1 -m_2)g-\dfrac{0.50 }{R} =\frac{1}{2}M a_y+(m_1 + m_2)a_y$$
$$ (m_1 -m_2)g-\dfrac{0.50 }{R} =\left[\frac{1}{2}M + m_1 + m_2 \right]a_y$$
Hence,
$$a_y=\dfrac{ (m_1 -m_2)g-\dfrac{0.50 }{R} }{\frac{1}{2}M + m_1 + m_2 }$$
Plugging the known;
$$a_y=\dfrac{ (4-2)9.8-\dfrac{0.50 }{0.06} }{\frac{1}{2}(2) + 4+ 2}=\bf 1.61\;\rm m/s^2$$
Now we can apply the kinematic equation on block 1.
$$\Delta y= v_{iy}t+\frac{1}{2}a_yt^2$$
where the block starts moving down from rest,
$$\Delta y=(0)t+\frac{1}{2}a_yt^2$$
Thus,
$$t=\sqrt{\dfrac{2\Delta y}{a_y}}$$
Plugging the known;
$$t=\sqrt{\dfrac{2(1)}{1.61}}=\color{red}{\bf 1.11}\;\rm s$$
