Since this is a relatively short string of DNA, we can count the number of mutations between the two strings. Firstly, Species A has thymine as the 2nd nucleotide, while Species B has adenine.2. Second, the 7th nucleotide of A has cytosine while B has thymine. Finally, the 16th nucleotide (3rd from the last) of A has adenine while B has thymine.
Since there are 3 point mutations, we multiply this by the frequency of 10 million years per mutation to get a time of 30 million years.