Question on MLE 
Can I do the part b in this way?I dont have any idea to handle part b.If it is wrong can anyone give me some hints to solve this problem.
 A: I think your solution for $(a)$ is correct i.e the mle for $\mu$ is $\bar{X}$ and mle for $\sigma^2$ is $\frac{\sum_{i=1}^{n} (X_i-\bar{X})^2}{n}$
Next you should know that $\frac{\sum_{i=1}^{n} (X_i-\bar{X})^2}{n}$ is an biased estimator of $\sigma^2$.
But $$\frac{\sum_{i=1}^{n} (X_i-\bar{X})^2}{n-1}$$ is an unbiased estimator of $\sigma^2$ i.e $$E[\frac{\sum_{i=1}^{n} (X_i-\bar{X})^2}{n-1}]=\sigma^2$$
You can see from above that $E[\sum_{i=1}^{n}(X_i-\bar{X})^2)]=(n-1)\sigma^2\tag{1}$
Next your already showed that:
$$\frac{\bar{X}}{\sqrt{\frac{\sum_{i=1}^{n}(X_i-\bar{X})^2}{n}}}$$ is the estimator for $\frac{\mu}{\sigma}$
You already know that $\bar{X}$ and $\sum_{i=1}^{n}(X_i-\bar{X})^2$ are independent.
So, $$E[\frac{\bar{X}}{\sqrt{\frac{\sum_{i=1}^{n}(X_i-\bar{X})^2}{n}}}]=E[\frac{\bar{X}}{\sqrt{\frac{\sum_{i=1}^{n} (X_i-\bar{X})^2(n-1)}{n(n-1)}}}]$$
And we know sample variance $$S^2=\frac{\sum_{i=1}^{n}(X_i-\bar{X})^2}{n-1}$$
$$E[\frac{\bar{X}}{\sqrt{\frac{\sum_{i=1}^{n}(X_i-\bar{X})^2}{n}}}]=E[\frac{\bar{X}}{\sqrt{\frac{\sum_{i=1}^{n} (X_i-\bar{X})^2(n-1)}{n(n-1)}}}]=\sqrt{\frac{1}{n-1}}E(\frac{\bar{X}}{S/\sqrt{n}})\tag{2}$$
And according to this paper $\frac{\bar{X}}{S/\sqrt{n}}$ has  a noncentral t-distribution with n-1 degrees of freedom and
non-centrality parameter $\frac{\mu\sqrt{n}}{\sigma}$
And read this paper we can see $E(\frac{\bar{X}}{S/\sqrt{n}})=\frac{\mu\sqrt{n}}{\sigma}$ put it into (2)
Finally, $$E[\frac{\bar{X}}{\sqrt{\frac{\sum_{i=1}^{n}(X_i-\bar{X})^2}{n}}}]=\sqrt{\frac{n}{n-1}}*\frac{\mu}{\sigma}$$
This is a biased estimator.
Interestingly,  this result is the same as my first try (treat $\frac{\sum_{i=1}^{n} (X_i-\bar{X})^2}{n}$ as a constant) if you still remember it. Is it a coincidence?
