**Solution:**

Option(**B**)
is correct

When we take successive powers of $2$ and find their remainders, we get the following cyclic patterns of cycle length $11$.

viz $2,4,8,16,32,64,39,78,67,45, 1$

i.e $2^{11}$ leaves a remainder $1$.

Thus $2^{89} = (2^{11})^8 (2)$ leaves a remainder of **2**.