1. Cathode reaction in dilute aqueous NaCl

In dilute aqueous NaCl, at the cathode water is reduced:

$2H_2O + 2e^- \rightarrow H_2 + 2OH^-$

So, to produce 1 mol of $H_2$, we need 2 mol of electrons.

2. Moles of electrons needed for $0.01$ mol $H_2$

Given:

$n(H_2)=0.01\text{ mol}$

From stoichiometry,

$n(e^-)=2\times 0.01 = 0.02\text{ mol}$

3. Charge required

Using:

$1\text{ Faraday} = 96500\text{ C mol}^{-1}$

Required charge:

$Q = n(e^-)\times F = 0.02\times 96500$

$Q=1930\text{ C}$

4. Use $Q = It$ to find time

Current passed:

$I=10\text{ milliampere}=10\times 10^{-3}=0.01\text{ A}$

Hence,

$t=\frac{Q}{I} = \frac{1930}{0.01}$

$t=193000\text{ s}$

$t = 1.93\times 10^5\text{ s} = 19.3\times 10^4\text{ s}$

5. Match with options

$19.3\times 10^4\text{ s}$ corresponds to Option B.

6. Verification with stored answer

Stored correct answer: B

Derived answer: B

They match.