Answer
$256$
Work Step by Step
W have to make 8 choices as the string length is 10 but the first and the last digits are already determined to be 1.
Each of the choice consists of choosing one of the 2 options 0 or 1.
Therefore, number of bit strings of length 10 which both begin and end with a one = $2\times2\times ... \times2$. [8 times]
=$2^8$
=256
