Find such a generating function for moments about the mean of the binomial distribution, and verify that the second derivative at $t=0$ is $n\theta (1-\theta)$.
Till now, I have found the $$M_{X-\mu}(t) = e^{-\mu t}{[1+\theta(e^t-1)]}^n$$ After derivating twice and putting $t=0$ I am not getting $n\theta (1-\theta)$. The generating function for moments about the mean that I have calculated wrong?
In that way you are getting the second moment of $X-\mu$ not of $X$.
The correct calculation is $M_X(t)=(1-\theta+\theta e^t)^n=n\theta(n\theta + 1- \theta)=\mathbb{E}[X^2]$.
Thus if you want $Var(X)=\mathbb{E}[X^2]-\mathbb{E}[X]^2=n\theta (1-\theta)$.