3 is CORRECT. Adding a front and back end zero to the number (0111011.100)2 results in (0011011.0100)2. Grouping the bits in groups of 4 about the radix point, interpret the number as follows: 0011 = 3; 1011 = A; Dot 1000 = 8. Therefore, (0111011.100)2 = (3B.8), using the 2^k conversion method.