Answer
No.
${\bf 1.4\times 10^{-27}}\;\rm m$
Work Step by Step
The uncertainty principle tells us that if the particle is confined to a certain region of space (in this case, the width of the hole), there is an uncertainty in its momentum, which corresponds to kinetic energy.
So we need to use the uncertainty principle to find the uncertainty in the momentum.
$$ \Delta x \Delta p_x \geq \frac{h}{2}$$
Solving for $ \Delta p_x $:
$$ \Delta p_x \geq \frac{h}{2 \Delta x} $$
Plug the known;
$$ \Delta p_x \geq \frac{6.63\times 10^{-34} }{2 \times 10 \times 10^{-6} } =\bf 3.315\times 10^{-29}\;\rm kg\cdot m/s \tag 1$$
Now we need to calculate the kinetic energy associated with $ \Delta p_x $
We know that the kinetic energy $ K $ is given by;
$$ K = \frac{(\Delta p_x )^2}{2m} $$
Plug the known;
$$ K = \frac{(3.315\times 10^{-29})^2}{2 (1.0 \times 10^{-16})} = \bf 5.5\times 10^{-42} \; \rm{J}\tag 2 $$
This is the kinetic energy the particle could potentially have due to the uncertainty in its momentum.
Now let's calculate the gravitational potential energy to Escape from the hole.
$$ U = m g h $$
Plug the known
$$ U = (1.0 \times 10^{-16}) ( 9.8) (1.0 \times 10^{-6}) =\bf 9.8 \times 10^{-22} \; \rm J \tag 3$$
From (2) and (3), it is obvious that the kinetic energy from the uncertainty principle is much less than the potential energy required to escape the hole.
Thus, the particle does not have enough energy to escape from the hole.
To find the deepest hole from which the particle could escape, we need to set the kinetic energy $ K $ equal to the gravitational potential energy $ U $ and then solve for $h$,
$$ K = U=mgh $$
where $h$ here is the height, NOT Planck's constant.
So,
$$ h=\dfrac{K}{mg} =\dfrac{\frac{1}{2}mv_x^2}{mg}$$
$$h=\dfrac{v_x^2}{2g} \tag 4$$
Now we need to find $v_x$ where we know that
$$\Delta v_x=\dfrac{\Delta p_x}{m}$$
Plug from (1);
$$\Delta v_x=\dfrac{(3.315\times 10^{-29})}{ (1.0 \times 10^{-16})}=\bf 3.315\times 10^{-13}\;\rm m/s$$
And since the average velocity is 0 m/s, then
$$-1.6575\times 10^{-13} \leq v_x\leq 1.6575\times 10^{-13} $$
So the maximum speed of the particle is $1.6575\times 10^{-13} $ m/s.
Plug that into (4) and plug the known;
$$h=\dfrac{(1.6575\times 10^{-13} )^2}{2(9.8)} $$
$$h=\color{red}{\bf 1.4\times 10^{-27}}\;\rm m$$
