In the given hypothetical fission reaction:

$$^{236}_{92} X \rightarrow \, ^{141}_{56} Y + \, ^{92}_{36} Z + 3 R$$

We need to determine the identity of particles denoted by $ R $. Let's use the conservation of charge and mass number (nucleon number) to identify $ R $.

First, for the conservation of nucleon number (mass number), we have:

$$236 = 141 + 92 + 3 \times A_R$$

Where $ A_R $ is the mass number of $ R $. This simplifies to:

$$236 = 233 + 3A_R$$

$$3A_R = 236 - 233$$

$$3A_R = 3$$

$$A_R = 1$$

Next, we use the conservation of charge (atomic number), we have:

$$92 = 56 + 36 + 3Z_R$$

Where $ Z_R $ is the atomic number of $ R $. This simplifies to:

$$92 = 92 + 3Z_R$$

$$3Z_R = 92 - 92$$

$$3Z_R = 0$$

$$Z_R = 0$$

Since the particle $ R $ has a mass number of 1 and atomic number of 0, it must be a neutron.

So, the identity of the emitted particles $ R $ is:

Option B: Neutron