1. Use Snell's law across parallel interfaces

For a stack of slabs with parallel interfaces, the quantity

$n\sin\theta$

remains constant at every refraction.

Here, the incident medium has refractive index

$n_0=1.6$

and the incident angle is

$\theta_0=30^\circ.$

So the invariant is

$n_0\sin\theta_0=1.6\times \sin 30^\circ=1.6\times \frac12=0.8.$

Thus, inside any slab of refractive index $n_m$,

$n_m\sin\theta_m=0.8.$

2. Condition for emergent ray to be parallel to the interface

The ray is said to be refracted out from the right side parallel to the interface between the $(m-1)^{\text{th}}$ and $m^{\text{th}}$ slabs.

That means, at the last refraction, the refracted ray makes angle

$90^\circ$

with the normal to that interface.

So at the exit surface, the refracted angle is $90^\circ$, i.e. the incidence inside the last slab is the critical angle.

Hence, for the last slab,

$n_m\sin\theta_m = 1\cdot \sin 90^\circ = 1.$

But from step 1, the constant value of $n\sin\theta$ is already

$0.8.$

So if the outside medium were air, this would be impossible. Therefore the interpretation is that the ray travels through slabs until it reaches a slab where the angle inside becomes $90^\circ$ relative to the normal to the horizontal interfaces, i.e.

$\sin\theta_m=1.$

Using the invariant,

$n_m\sin\theta_m=0.8 \implies n_m(1)=0.8.$

So the required slab has refractive index

$n_m=0.8.$

3. Use the given refractive index pattern

The slab indices are given by

$n_m = n - m\Delta n$

with

$n=1.6,\qquad \Delta n=0.1.$

So,

$n_m=1.6-0.1m.$

Set this equal to $0.8$:

$1.6-0.1m=0.8.$

Therefore,

$0.1m=0.8$

$m=8.$

4. Final answer

$\boxed{8}$

This matches the stored correct answer.