10 cc = 10 mL

10 mL x 0.2 U/mL = 2 U

2 U/(5 U/mg) = 0.4 mg

Since it is difficult to accurately weigh 0.4 mg of enzyme, you should start by making 1 mL of a concentrated stock solution, let's say 100X, i.e., 20 U/mL

1 mL x 20 U/mL / (5 U/mg) = 4 mg

If you have an analytical balance capable of weighing 1 mg accurately, you could weigh out 4 mg of enzyme and dissolve it in a storage buffer, then use this 100X stock solution to prepare your 0.2 U/mL solution.

0.1 mL of 20 U/mL + 9.9 mL of buffer = 10 mL of 0.2 U/mL

If you don't have an analytical balance handy, it may be easier to dissolve the entire contents of the enzyme sample to avoid having to weigh it out. In that case, you may want to make an even more concentrated stock solution. The stock solution can be flash-frozen in a dry ice ethanol bath or liquid nitrogen in convenient-sized aliquots for storage in an ultralow freezer.

Thank you so much

What Adam Shapiro showed you is all correct. Please however understand that you stated 5 Units/mg Protein. Is the Amylase material you are using a solid and is the % Protein 100%, if so what Adam described is correct. However, if it is not 100% protein you will have to calculate the Units/mg solid from the fact you have a % protein and a Units/mg Protein value. Then determine how much mg solid you need to weigh to dissolve as Adam described