Electrolysis of NaCl is

$$\mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}(\mathrm{aq}) \rightarrow \mathrm{NaOH}(\mathrm{aq})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g})+\frac{1}{2} \mathrm{H}_2(\mathrm{~g})$$

Since during electrolysis pH changes to 12

So $\left[\mathrm{OH}^{\ominus}\right]=10^{-2}$ and $\left[\mathrm{H}^{+}\right]=10^{-12}$

So by Faraday law

Gram amount of substance deposited $=$ Amount of electricity passed

$$\begin{aligned} & 10^{-2} \times \frac{600}{1000} \times 96500=\mathrm{I} \times \mathrm{t} \\ & \frac{10^{-2} \times 600}{1000} \times 96500=\mathrm{I} \times 5 \times 60 \\ & \mathrm{I}=\frac{10^{-2} \times 600 \times 96500}{1000 \times 5 \times 60} \\ & \mathrm{I}=1.93 \text { ampere } \end{aligned}$$

So, $\mathrm{I = 2}$ ampere (nearest integer)