1. Write the half-cell reaction for hydrogen electrode

For a hydrogen electrode, $2H^+ + 2e^- \rightleftharpoons H_2(g)$

2. Use the Nernst equation

For the reduction reaction, $E = E^\circ - \frac{0.059}{2} \log \frac{P_{H_2}}{[H^+]^2}$

For the standard hydrogen electrode, $E^\circ = 0$

Assuming hydrogen gas is at standard pressure, $P_{H_2}=1$ atm. So, $E = -\frac{0.059}{2} \log \frac{1}{[H^+]^2}$

This simplifies to $E = -\frac{0.059}{2}(-2\log[H^+])$ $E = 0.059\log[H^+]$

Since $\mathrm{pH} = -\log[H^+]$ we get $\log[H^+] = -\mathrm{pH}$

Therefore, $E = -0.059\times \mathrm{pH}$

3. Substitute pH = 3

$E = -0.059\times 3 = -0.177\,\text{V}$

4. Express in the required form

$-0.177\,\text{V} = -17.7\times 10^{-2}\,\text{V}$

As an integer-type answer, this is taken as $18\times 10^{-2}\,\text{V}$ with negative sign for the electrode potential.

So the potential is approximately $-18\times 10^{-2}\,\text{V}$

Hence the blank should be filled with 18.