Given,

Current passed a 10 mA.

Moles of Hydrogen released are 0.01 moles.

In the electrolysis of aqueous solution sodium chloride, as the reduction potential of sodium ion is less than the hydrogen ion, the reaction that takes place at the cathode could be given as,

2H$$^+$$ + 2e$$^-$$ $$\to$$ H$$_2$$

0.01 moles of hydrogen would require 2 $$\times$$ 0.01 = 0.02 moles of electrons (or) 0.02 Faradays.

We know the formula to calculate the amount of Faradays passed.

Amount of Faraday's passed = $$\frac{I(A)\times t(s)}{96500~\mathrm{C~mol^{-1}}}$$

On rearranging the equation to calculate the time taken we get,

t = Amount of Faraday's passed $$\times$$ 96500 $$\times$$ I

$$0.02 = {{{{10\,mA} \over {1000\,mA/A}} \times t(s)} \over {96500\,C\,mo{l^{ - 1}}}}$$

$$\Rightarrow~t = 0.02\times96500\times0.01$$

On multiplying we get,

$$\Rightarrow~t=19.3\times10^4$$ s

The time required to liberate 0.01 moles of H$$_2$$ gas at the cathode is 19.3 $$\times$$ 10$$^4$$ s.