$E[\frac{\sum \epsilon_i}{\sum x_i}]=0$.
To see this, note that $\sum x_i$ is non-random and so the constant $\frac{1}{\sum x_i}$ can be taken outside of the expectation. Mean while, the linearity of expectation tells us that:
$E[\sum \epsilon_i] = \sum E[\epsilon_i] = \sum 0 = 0$ since each $\epsilon_i \sim N(0,1)$ so that $E[\epsilon_i] = 0 \forall i$
Putting this together gives:
$E[\theta^*] = \theta + E[\frac{\sum \epsilon_i}{\sum x_i}]= \theta + \frac{1}{\sum x_i} E[\sum \epsilon_i] = \theta + 0 = \theta$
Edit: Due to the odd wording of the question, suppose that $x_i$ are indeed random and not yet sampled. So that $y_i = \theta x_i + \epsilon_i$ with both $x_i$ and $\epsilon_i$ being standard normal random variables. Then the above solution still works provided we can assume $x_i$ and $\epsilon_i$ are independent.
This, in my opinion, is a completely valid assumption since $\epsilon_i$ are just stated to be random noise and so should not be dependent on the covariates $x_i$.
If this is the case, then we can use the fact that for independent random variables $X$,$Y$, and functions $f(x)$,$g(y)$, we know that:
$E[f(X)g(Y)] = E[f(X)]\times E[g(Y)]$
So, in this context $f(x) = \frac{1}{\sum x_i}$ and $g(\epsilon) = \sum \epsilon_i$
This gives us that:
$E[\frac{\sum \epsilon_i}{\sum x_i}] = E[\frac{1}{\sum x_i}]\times E[\sum \epsilon_i] = E[\frac{1}{\sum x_i}] \times 0 = 0$
So that we don't actually need to evaluate the expectation of $\frac{1}{\sum x_i}$.
If on the other hand we can't assume they are independent, I don't think there's anything you can do.
