Discussion
There is no subfield with k elements to the field with 64 elements if k is :
Answer: A
There is no subfield with 16 elements.
The field with 64 elements is \(\mathbf{F}_{2^{6}}\). Its subfields are exactly the fields \(\mathbf{F}_{2^{d}}\) with d dividing 6, that is \(d = 1, 2, 3, 6\), giving subfields of 2, 4, 8 and 64 elements.
16 = \(2^{4}\), and 4 does not divide 6, so \(\mathbf{F}_{16}\) is not a subfield. (Its multiplicative group would have order 15, which does not divide 63.)
The fields with 2, 4 and 8 elements all occur, so those options are wrong.
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