If I am not mistaken, you have one fixed base part, to which the torque creating motor is attached, and 3 moving parts of the mechanism (the three drawer slides). This mechanism has one degree of freedom. You seem to be ignoring the masses of the three slides, so I would ignore them too. It is not clear to me whether you are lifting the load vertically with this mechanism or horizontally (i.e. presence of gravity or no presence of gravity).
I tried to apply Lagrangian formalism to this model, and in particular D'Alembert's principle.
Observe that by construction of the mechanism, the belt on the fixed base part is the one driven by the motor, so when in motion, we can introduce displacement $x$ of the said belt. That displacement displaces the first slide at distance $x$ from its initial position, the second slide is displaced at distance $2x$ from its initial position, and the third slide is displaced at distance $3x$ from its initial position. Since the load is on the third slide, the displacement of the load is $x_{load} = 3x$.
It seems to me that, since the slides are massless, the Lagrangian of this system should be
$$L \, =\, \frac{m}{2} \left(\frac{dx_{load}}{dt}\right)^2 \, -\, m\, g\, x_{load}$$ (simply the kinetic minus the potential energy of the load, as the load is the only object with mass). Now, using the displacement $x$ as a generalized coordinate, the Lagrangian becomes
$$L \, =\, \frac{m}{2} \left(3\, \frac{dx}{dt}\right)^2 \, -\, m\, g\, \big(3\,x\big) \, =\, \frac{9\,m}{2} \left( \frac{dx}{dt}\right)^2 \, -\,3\, m\, g\, x$$
Since the torque $\tau$ in question generates a linear force $F$ along the belt on the fixed base part, the said force causes virtual displacement $\delta x$ of the belt. D'Alembert's principle then implies
$$\left(\frac{\delta}{\delta x} L\right) \delta x \,= \, F\delta x$$ which holds if and only if
$$\frac{\delta}{\delta x} L \,= \, F$$
From Lagrangian mechanics,
$$\frac{\delta}{\delta x} L \, =\, \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} \, -\, \frac{\partial L}{\partial {x}} \, = \, F$$
($\dot{x} = \frac{dx}{dt}$) For this model, the equation of motion is therefore
$$ {9\,m} \frac{d^2x}{dt^2}\, + \,3\, m\, g \, =\, F$$
Now, the linear force $F$ is related to the torque by the relation $$\tau \, =\, r \, F$$ Thus, finally we get
$$ {9\,m} \frac{d^2x}{dt^2}\, + \,3\, m\, g \, =\, \frac{\tau}{r}$$ or alternatively
$$ \frac{d^2x}{dt^2}\, \, =\, \frac{\tau}{9\,m\,r} \, -\, \,\frac{g}{3}$$
Assuming that the torque $\tau$ is constant for half the time of the full extension of the mechanism, we can integrate the latter equation twice, obtaining the law of motion
$$x \, =\, \frac{t^2}{2}\left(\frac{\tau}{9\,m\,r} \, -\, \,\frac{g}{3}\right)$$
Hence, the actual law of motion of the load is
$$x_{load} \, =\, \frac{t^2}{2}\left(\frac{\tau}{3\,m\,r} \, -\, \,{g}\right)$$
If the full extension of the mechanism is $x_{load} \,=\, D$ and the time to achieve it is $T$, then one simple scenario is uniformly accelerating, applying constant toque $\tau$ in one direction , for half of the total time $T$, then uniformly decelerating, applying constant toque $\tau$ in the other direction, for the remaining half of the total time $T$. This means that we want when $t = \frac{T}{2}$ the displacement to be $x_{load} = \frac{D}{2}$. So plug it
$$\frac{D}{2} \, =\, \frac{(T/2)^2}{2}\left(\frac{\tau}{3\,m\,r} \, -\, \,{g}\right)$$
$${D} \, =\, \frac{T^2}{4}\left(\frac{\tau}{3\,m\,r} \, -\, \,{g}\right)$$
and if you solve for $\tau$ you get
$$\tau \,=\, \frac{12\, m\, r\,D}{T^2} \, + \, 3\, m\, r\,g$$
So maybe you would want a torque at leas greater than this, i.e.
$$\tau \,\geq\, \frac{12\, m\, r\,D}{T^2} \, + \, 3\, m\, r\,g$$
If gravity doesn't play a role, just turn it of by setting $g=0$
$$\tau \,\geq\, \frac{12\, m\, r\,D}{T^2} $$
