Write the half-reaction and standard potential

The given half-cell is: $2H^+_{(aq)} + 2e^- \rightarrow H_2(g)$

This is the standard hydrogen electrode, so $E^\circ = 0$

Apply the Nernst equation

For the reduction reaction, $E = E^\circ - \frac{0.06}{n}\log Q$ at $298\,\text{K}$.

Here, $n=2$.

Write the reaction quotient

For $2H^+ + 2e^- \rightarrow H_2$ the reaction quotient is $Q = \frac{P_{H_2}}{[H^+]^2}$

Given: $[H^+] = 1\,\text{M}, \quad P_{H_2}=2\,\text{atm}$

So, $Q = \frac{2}{(1)^2} = 2$

Substitute into Nernst equation

$E = 0 - \frac{0.06}{2}\log 2$

Using $\log 2 = 0.3$, $E = -0.03 \times 0.3$ $E = -0.009\,\text{V}$

Express in the required form

We need: $(-)\;\underline{\hspace{2cm}} \times 10^{-2}\,\text{V}$

Since $-0.009\,\text{V} = -0.9 \times 10^{-2}\,\text{V}$

The integer to be filled is: $\boxed{1}$ (taking the nearest integer)

Comparison with stored answer

Derived answer = $1$

Stored correct answer = $1$

Hence, they agree.