Step-by-Step Solution:

1. Identify the reactants and products.

   • The reactant is rhombic sulphur, which has the molecular formula $S_8$. In its elemental form, the oxidation state of sulphur is 0.
   • The oxidizing agent is concentrated nitric acid, $HNO_3$. The oxidation state of Nitrogen in $HNO_3$ is +5.
   • The problem states that sulphur is oxidized to a compound with its highest oxidation state. The highest oxidation state of sulphur (a group 16 element) is +6. The common stable compound formed is sulphuric acid, $H_2SO_4$.
   • When concentrated $HNO_3$ acts as an oxidizing agent, it is typically reduced to nitrogen dioxide, $NO_2$, where the oxidation state of Nitrogen is +4.
   • Water ($H_2O$) is also formed in the reaction.

2. Write the unbalanced chemical equation. The unbalanced equation based on the identified species is: $S_8 + HNO_3 \rightarrow H_2SO_4 + NO_2 + H_2O$

3. Balance the equation using the oxidation number method.

   • Oxidation half-reaction: Sulphur is oxidized from 0 to +6.

     • Change in oxidation state per S atom = $+6 - 0 = +6$.
     • Since one molecule of rhombic sulphur contains 8 atoms ($S_8$), the total increase in oxidation number for one mole of $S_8$ is $8 \times (+6) = +48$.

   • Reduction half-reaction: Nitrogen is reduced from +5 in $HNO_3$ to +4 in $NO_2$.

     • Change in oxidation state per N atom = $+4 - (+5) = -1$.

   • Balance the change in oxidation states: To make the total increase in oxidation number equal to the total decrease, we need to multiply the reduction half-reaction by 48. This gives the stoichiometric ratio between $S_8$ and $HNO_3$ as 1:48.

   • This also implies that 1 mole of $S_8$ produces 8 moles of $H_2SO_4$ and 48 moles of $NO_2$.

4. Write the partially balanced equation. $1 S_8 + 48 HNO_3 \rightarrow 8 H_2SO_4 + 48 NO_2 + x H_2O$ We need to find the coefficient 'x' for water by balancing the remaining atoms (Hydrogen and Oxygen).

5. Balance the Hydrogen and Oxygen atoms.

   • Balance Hydrogen (H) atoms:

     • On the left-hand side (LHS): $48 \times 1 = 48$ H atoms.
     • On the right-hand side (RHS): $(8 \times 2) + (x \times 2) = 16 + 2x$ H atoms.
     • Equating H atoms on both sides: $48 = 16 + 2x$
     • $2x = 48 - 16 = 32$
     • $x = 16$

   • The balanced chemical equation is: $S_8 + 48 HNO_3 \rightarrow 8 H_2SO_4 + 48 NO_2 + 16 H_2O$

   • (Check) Balance Oxygen (O) atoms:

     • LHS: $48 \times 3 = 144$ O atoms.
     • RHS: $(8 \times 4) + (48 \times 2) + (16 \times 1) = 32 + 96 + 16 = 144$ O atoms.
     • The equation is correctly balanced.

6. Calculate the amount of water produced.

   • The question asks for the amount of water produced from the oxidation of 1 mole of rhombic sulphur ($S_8$).
   • From the balanced equation, the stoichiometric relationship is: 1 mole of $S_8$ produces 16 moles of $H_2O$.
   • Moles of water produced = 16 mol.

7. Convert moles of water to mass in grams.

   • Given: Molar mass of water = 18 g mol$^{-1}$.
   • Mass of water = Moles of water $\times$ Molar mass of water
   • Mass of water = $16 \text{ mol} \times 18 \text{ g mol}^{-1} = 288 \text{ g}$.

The amount of water produced is 288 g.