1. Identify the half-cells

The cell is:

$\text{Pt} | H_2(g,1\,\text{bar}) | HCl(aq) | AgCl(s) | Ag(s) | \text{Pt}$

This consists of:

• Left electrode: hydrogen electrode
• Right electrode: silver-silver chloride electrode

So,

$E_{\text{cell}} = E_{\text{right}} - E_{\text{left}}$

Given:

$E_{\text{cell}} = 0.92\,\text{V}$

and concentration of HCl is $10^{-6}$ molal, so for this dilute solution we take

$[H^+] = [Cl^-] = 10^{-6}$

2. Potential of hydrogen electrode

For the hydrogen electrode reaction

$2H^+ + 2e^- \rightarrow H_2(g)$

Nernst equation:

$E_{H^+/H_2} = E^\circ - \frac{0.06}{2} \log \frac{P_{H_2}}{[H^+]^2}$

Since $E^\circ=0$ and $P_{H_2}=1$ bar,

$E_{H^+/H_2} = -\frac{0.06}{2} \log \frac{1}{(10^{-6})^2}$

$= -0.03 \log (10^{12})$

$= -0.03 \times 12 = -0.36\,\text{V}$

So,

$E_{\text{left}} = -0.36\,\text{V}$

3. Potential of AgCl/Ag electrode

The electrode reaction is

$AgCl(s) + e^- \rightarrow Ag(s) + Cl^-$

For this electrode,

$E = E^\circ - 0.06 \log [Cl^-]$

Since $[Cl^-] = 10^{-6}$,

$E_{\text{right}} = E^\circ - 0.06 \log(10^{-6})$

$= E^\circ - 0.06(-6)$

$= E^\circ + 0.36$

4. Use cell potential

Now,

$E_{\text{cell}} = E_{\text{right}} - E_{\text{left}}$

$0.92 = (E^\circ + 0.36) - (-0.36)$

$0.92 = E^\circ + 0.72$

$E^\circ = 0.92 - 0.72 = 0.20\,\text{V}$

5. Final answer

Therefore, the standard electrode potential of the $(AgCl/Ag,Cl^-)$ electrode is

$\boxed{0.20\,\text{V}}$

So the correct option is D.