Write the oxidation half-reaction of water

In acidic medium, oxidation of water to oxygen is: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$

Relate moles of water to moles of electrons

From the reaction, $2\text{ mol } H_2O \rightarrow 4\text{ mol } e^-$

Therefore, $1\text{ mol } H_2O \rightarrow 2\text{ mol } e^-$

Use Faraday's constant

Charge required for 1 mole of electrons is: $1F = 96500\,C$

So for $2$ moles of electrons: $Q = 2F = 2 \times 96500 = 193000\,C$

Express in the required form

$193000\,C = 1.93 \times 10^5\,C$

Since the blank is in the form $\underline{\hspace{2cm}} \times 10^5\,C$, the value is approximately: $2 \times 10^5\,C$

Compare with stored answer

Derived answer = $2$

Stored correct answer = $2$

Hence, they agree.