At STP, the volume of $H_2$ gas produced is $0.112 \text{ L}$. The number of moles of $H_2$ is: $$ \begin{aligned} n &= \frac{0.112 \text{ L}}{22.4 \text{ L/mol}} = 0.005 \text{ mol} \end{aligned} $$ The cathode reaction for the electrolysis of $NaCl$ solution is: $$ 2H^+(aq) + 2e^- \rightarrow H_2(g) $$ From the reaction, $1 \text{ mol } H_2$ requires $2 \text{ mol } e^-$. So, $0.005 \text{ mol } H_2$ requires: $$ \begin{aligned} Q &= 0.005 \times 2 \times 96500 \text{ C} = 965 \text{ C} \end{aligned} $$ Given time $t = 1 \text{ hour} = 3600 \text{ seconds}$. Using the formula $Q = I \times t$: $$ \begin{aligned} I &= \frac{Q}{t} = \frac{965}{3600} \text{ A} \approx 0.268 \text{ A} \end{aligned} $$ Thus, the required current is $0.268 \text{ A}$.