When intermolecular forces vanish, water vapour behaves as an ideal gas. We can calculate the volume using the Ideal Gas Law or the molar volume at STP.

1. Calculate Moles ($n$): Molar mass of water ($H_2O$) = $18 \text{ g/mol}$. Given mass = $4.5 \text{ kg} = 4500 \text{ g}$. $n = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{4500}{18} = 250 \text{ mol}$.

2. Calculate Volume ($V$): At STP (Standard Temperature and Pressure), one mole of an ideal gas occupies $22.4 \text{ Litres}$ (standard approximation for NEET calculations; older definition). $V = n \times 22.4 \text{ L/mol}$ $V = 250 \times 22.4 = 5600 \text{ Litres}$.

3. Convert to SI Units ($m^3$): Since $1 \text{ m}^3 = 1000 \text{ Litres}$: $V = \frac{5600}{1000} \text{ m}^3 = 5.6 \text{ m}^3$.

(Note: Using the modern IUPAC STP molar volume of $22.7 \text{ L}$ would yield $5.675 \text{ m}^3$, which is closest to option A).