Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
First, we need to find the uncertainty in the momentum and velocity of the dust particle. So we need to use the Heisenberg uncertainty principle.
$$ \Delta p_x\Delta x=\dfrac{h}{2}$$
$$ \Delta p_x = \frac{h}{2 \Delta x}$$
Plug the known
$$ \Delta p_x = \frac{6.63 \times 10^{-34} }{2 (1.0 \times 10^{-6} ) } = 3.315 \times 10^{-28} \; \rm{kg\cdot m/s}
$$
From this, the uncertainty in velocity is:
$$
\Delta v_x = \frac{\Delta p_x}{m} = \frac{3.315 \times 10^{-28}}{1.0 \times 10^{-15}} = 3.32 \times 10^{-13} \; \rm{m/s}
$$
The velocity uncertainty is symmetric, so the range of velocities is
$$ - 1.66 \times 10^{-13} \; {\rm m/s}\leq v_x\leq 1.66 \times 10^{-13}\rm \;m/s$$
Now, we need to determine the time it takes for the dust particle to fall 1 meter to the detection circle.
We can use the kinematic formula of
$$y_f-y_i=v_{iy}t+\frac{1}{2}a_yt^2$$
where $v_{iy}=0$ m/s since it falls from rest, and $a_y=-g$
$$0-1=0+\frac{-1}{2}gt^2$$
Thus,
$$t=\sqrt{\dfrac{2}{g}}=\sqrt{\dfrac{2}{9.8}}=\bf 0.452\;\rm s$$
Now we need to find the horizontal distance traveled by the dust particle through this time interval of 0.452 s.
$$\Delta x=v_x t$$
Plug from the velocity range;
$$\Delta x=( 1.66 \times 10^{-13})(0.452)=\bf 7.50\times 10^{-14}\;\rm m$$
This horizontal spread (due to uncertainty in momentum) is extremely small and not detectable, as it is much smaller than the 1.0 $\mu$m diameter of the hole.
$$\color{blue}{\bf [b]}$$
To get a detection circle diameter of 1.1 $\mu$m, the dust particle needs to move by 0.05 $\mu$m in each direction (to increase the detection circle radius from 0.50 $\mu$m to 0.55 $\mu$m).
The time required to achieve this horizontal displacement under gravitational acceleration is then
$$ t =\dfrac{\Delta x}{v_x}= \frac{0.05\times 10^{-6}}{1.66 \times 10^{-13} } \approx 3.01 \times 10^5 \; \rm{s} $$
This is a huge time but let's complete and find the height $d$ needed for this increase in time.
$$ y_f - y_i = \frac{1}{2} g t^2 $$
$$ d= \frac{1}{2}(9.8)(3.01 \times 10^5)^2 =\color{red}{\bf 4.44\times 10^{11} }\; \rm{m} $$
This distance is significantly larger than any practical distance, even exceeding the diameter of Earth's orbit around the sun. Therefore, to observe such a small diffraction effect that increases the detection diameter by 10%, the detector would need to be positioned at an impractically large distance.
