If the genetic code consisted of four bases per codon instead of three bases per codon, then what will be the possible number of codons that code for 20 types of amino acids? (Given the stops codons are same)

The correct option is D 253

If genetic code consisted of four bases per codon, then with the four nucleotides( A,T, G, C) available number of codons possible will be ($4^4$) =256
It is given in the question that stop codons are same so, 256-3(stop codons) = 253
Thus, 253 will be the possible number of codons coding for 20 different amino acids provided the codons are tetraplet.