$$\hat\theta_g=\frac{1}{n}\sum_{i=1}^{n}\frac{e^2}{2}e^{-2X_i}$$
where the $X_i$ are an iid sample from a pdf.
Show that $$Var(\hat\theta_g)=\frac{1}{n}Var(\frac{e^2}{2}e^{-2X_1})$$
My attempt:
$$Var(\hat\theta_g)=Var(\frac{1}{n}\sum_{i=1}^{n}\frac{e^2}{2}e^{-2X_i})$$
Since $X_i$ are an iid sample
$$Var(\hat\theta_g)=Var(\frac{1}{n}\sum_{i=1}^{n}\frac{e^2}{2}e^{-2X_1})$$
$$Var(\hat\theta_g)=Var(\frac{1}{n}n\frac{e^2}{2}e^{-2X_1})$$
$$Var(\hat\theta_g)=Var(\frac{e^2}{2}e^{-2X_1})$$
But the answer is:
$$Var(\hat\theta_g)=\frac{1}{n}Var(\frac{e^2}{2}e^{-2X_1})$$
If $(\xi_i)$ is i.i.d., then $\mathrm{Var}\left(\frac1n\sum\limits_{i=1}^n\xi_i\right)=\frac1n\mathrm{Var}\left(\xi_1\right)$, not $\mathrm{Var}\left(\frac1nn\xi_1\right)$. The rest of your computations is kosher.
Recall that, for every scalar $a$, $\mathrm{Var}\left(a\xi\right)=a^2\mathrm{Var}\left(\xi\right)$ and that, for every independent $\xi$ and $\eta$, $\mathrm{Var}\left(\xi+\eta\right)=\mathrm{Var}\left(\xi\right)+\mathrm{Var}\left(\eta\right)$.
In your setting, choose $\xi_i=\frac12\mathrm e^2\mathrm e^{-2X_i}$ for every $i$, apply $\mathrm{Var}\left(\frac1n\sum\limits_{i=1}^n\xi_i\right)=\frac1{n^2}\mathrm{Var}\left(\sum\limits_{i=1}^n\xi_i\right)$, then $\mathrm{Var}\left(\sum\limits_{i=1}^n\xi_i\right)=\sum\limits_{i=1}^n\mathrm{Var}\left(\xi_i\right)$ because the random variables $(\xi_i)$ are independent, and finally $\mathrm{Var}\left(\xi_i\right)=\mathrm{Var}\left(\xi_1\right)$ for every $i$ because the random variables $(\xi_i)$ are identically distributed.