Glycine in aqueous neutral solution (i.e. add water to glycine) occurs as zwitter ion:
![enter image description here](https://cdn.statically.io/img/i.sstatic.net/1G8k5.png)
The $\mathrm{p}K_\mathrm{a}$ of the ammonium group is 9.6. The $\mathrm{p}K_\mathrm{a}$ of the carboxylic acid is 2.3 (quite different from acetic acid, which is around 4.7). Given that the carboxylic acid group is a stronger acid than the ammonium group, the zwitter ion forms, as shown above.
Adding HCl to the zwitter ion will protonate the carboxylate, so the $\mathrm{p}K_\mathrm{a}$ of 2.3 is relevant (the ammonium is already protonated, and remains protonated as you add strong acid). In the problem, you are adding just enough HCl to protonate half of the carboxylate. The result is a 1:1 buffer, a mixture of the zwitter ion (net neutral) and the twice protonated species (net charge of +1).
With this information, you should be able to plug in concentrations and $\mathrm{p}K_\mathrm{a}$ into the buffer equation (the Henderson-Hasselbalch equation).