(B) is the correct choice. 
We note that there are 3 consonants and 3 vowels E, A and O. Since no two vowels have to be together, the possible choice for vowels are the places marked as ‘X’. 
X M X C X T X, these vowels can be arranged in 4P3 ways 3 consonants can be arranged in 3 ways. 
Hence, the required number of ways = 3! × 4P3 = 144