1. Identify reagent $Y$ from the given tests

   The solution of $Y$ gives:

   • with potassium hexacyanoferrate(II), an intense blue precipitate that dissolves in excess reagent,
   • with potassium hexacyanoferrate(III), a brown coloration due to formation of $Z$.

   These are characteristic tests of $Fe^{3+}$.

   In salt analysis:

   • $Fe^{3+}$ with potassium ferrocyanide, $K_4[Fe(CN)_6]$, gives Prussian blue: $$$$

ightarrow Fe_4[Fe(CN)_6]_3$$

• $Fe^{3+}$ with potassium ferricyanide, $K_3[Fe(CN)_6]$, gives a brown coloration due to ferric ferricyanide: $$$$

ightarrow Fe[Fe(CN)_6]$$

2. Identify $X$ from methylene blue formation

   In strongly acidic medium, $p$-amino-$N,N$-dimethylaniline reacts in presence of sulfide / hydrogen sulfide and then with $Fe^{3+}$ to form methylene blue.

   So:

   • $X$ is sulfide-containing species,
   • $Y$ is $Fe^{3+}$.

   This confirms our identification of $Y$.

3. Determine $Z$

   Since $Y = Fe^{3+}$, treatment with potassium hexacyanoferrate(III), i.e. $K_3[Fe(CN)_6]$, gives the brown compound:

   $$$$

ightarrow Fe[Fe(CN)_6]$$

Therefore, $Z = Fe[Fe(CN)_6]$

4. Check options

   • A: $Mg_2[Fe(CN)_6]$ — magnesium salt, not relevant.
   • B: $Fe[Fe(CN)_6]$ — matches the brown ferric ferricyanide.
   • C: $Fe_4[Fe(CN)_6]_3$ — Prussian blue, formed with ferrocyanide, not ferricyanide.
   • D: $K_2Zn_3[Fe(CN)_6]_2$ — zinc salt, not relevant.

5. Final answer

   The compound $Z$ is: $\boxed{Fe[Fe(CN)_6]}$ Hence, the correct option is B.