1. Write the reaction of sodium with water

$2Na + 2H_2O \rightarrow 2NaOH + H_2$

From the equation, 1 mole of Na produces 1 mole of NaOH.

2. Calculate moles of sodium used

Given mass of sodium = $0.69\,\text{g}$

Molar mass of sodium = $23\,\text{g mol}^{-1}$

$n(\text{Na}) = \frac{0.69}{23} = 0.03\,\text{mol}$

Therefore,

$n(\text{NaOH}) = 0.03\,\text{mol}$

3. Neutralisation reaction with HCl

$NaOH + HCl \rightarrow NaCl + H_2O$

This shows that 1 mole of NaOH reacts with 1 mole of HCl.

So, moles of HCl required =

$n(\text{HCl}) = 0.03\,\text{mol}$

4. Find concentration of HCl solution

Given HCl contains $73\,\text{g L}^{-1}$.

Molar mass of HCl =

$1 + 35.5 = 36.5\,\text{g mol}^{-1}$

So molarity of HCl is

$M = \frac{73}{36.5} = 2\,\text{mol L}^{-1}$

5. Calculate volume of HCl needed

Using

$V = \frac{n}{M}$

$V = \frac{0.03}{2} = 0.015\,\text{L}$

Convert to mL:

$0.015\,\text{L} = 15\,\text{mL}$

6. Final Answer

Required volume of HCl =

$\boxed{15\,\text{mL}}$