In an undirected random graph of 11 vertices the probability of an edge being present between a pair of vertices is 2/5. What is the expected number of unordered cycle of length 3. I think the answer should be
$${11\choose 3} \cdot\left(\frac 25\right)^3=10.56\approx 10.$$
But the options available are 22, 33, 44, 55. please help me
Your answer is correct. There are $\binom{11}3$ potential unordered cycles, and each of them exists with probability $\left(\frac25\right)^3$. These probabilities aren't independent, but they don't have to be in order to apply the linearity of expectation. There must be either an error in the text or a misunderstanding of the problem statement.