Alkaline glycine buffer molarity
As per comments and revisiting the question and my first answer, I am amending my answer to reflect the correct one. 

Because you are converting glycine to its sodium salt, the final solution concentration will be 2 times the concentration of glycine.

How many moles of glycine are in $25\,\mathrm{ml}$ of stock solution?
$${0.2\,\mathrm{mol}\,\ce{glycine}\over 1000\,\mathrm{ml}} = {x\,\mathrm{mol}\,\ce{glycine}\over 25\,\mathrm{ml}}\implies x = 0.005\,\mathrm{mol}\,\ce{glycine}$$

You need $0.005\,\mathrm{mol}$ of $\ce{NaOH}$ to convert the glycine to the sodium salt, so:

Total amount of substance in buffer solution:
$$0.005\,\mathrm{mol} + 0.005\,\mathrm{mol} = 0.01\,\mathrm{mol}$$

Total volume of buffer solution:
$$100\,\mathrm{ml} = 0.1\,\mathrm{L}$$

Molarity of buffer solution:
$${0.01\,\mathrm{mol}\over 0.1\,\mathrm{L}} = 0.1\,\mathrm{M}$$