Since we are looking at a sample mean we have that
$$\bar X_n \sim_{approx} N \left(\theta ,\frac{\sigma^2}{n} \right)$$
which holds for "large but finite $n$" -since if $n\rightarrow \infty$ the sample mean unscaled does not have a distribution, since it becomes a constant, as the comments noted. Then our conclusions will be approximate (and we keep $n$ fixed).
We have
$$Z= \left[ \frac{\left(\bar X_n -\theta \right) \sqrt{n}} {\sigma} \right]^2 \sim_{approx} \chi^2 \left({1} \right) $$
Expand the functional form for $Z$:
$$Z= \frac n{\sigma^2} \Big (\bar X_n^2 -2\theta \bar X_n + \theta^2\Big) $$
$$\Rightarrow \bar X_n^2 =\frac {\sigma^2}{n}Z + 2\theta \bar X_n - \theta^2$$
We know the moments of $\bar X_n$, and we know the moments of $Z$, so we don't really need the distribution of $\bar X_n^2$. I guess the OP can take it up from here. One should not forget to consider what may be the covariance of $Z$ and $\bar X_n$.
ADDEDNDUM
(I am completing this answer after a I gave the OP time to work it through).
We want to consider the estimator for $\theta^2$, denote it $\hat h_n$,
$$\hat h_n = \bar X_n^2- \frac{\sigma^2}{n}$$
UNBIASEDNESS
Its expected value is
$$ E(\hat h_n) = E\left(\bar X_n^2- \frac{\sigma^2}{n}\right) = E\left(\frac {\sigma^2}{n}Z + 2\theta \bar X_n - \theta^2- \frac{\sigma^2}{n}\right)$$
Since $E(Z) = 1$ and $E(\bar X) = \theta$ we obtain
$$E(\hat h_n) = \frac {\sigma^2}{n} + 2\theta^2 -\theta^2 - \frac {\sigma^2}{n} = \theta^2$$
So $\hat h_n$ is unbiased.
EFFICIENCY
The varinace of the estimator is
$$\operatorname{Var}(\hat h_n) = \operatorname{Var}\left(\frac {\sigma^2}{n}Z + 2\theta \bar X_n - \theta^2- \frac{\sigma^2}{n}\right) $$
$$= \left(\frac {\sigma^2}{n}\right)^2\operatorname{Var}(Z) + \left(2\theta\right)^2\operatorname{Var}(\bar X_n) +2\frac {\sigma^2}{n}2\theta\operatorname{Cov}(Z,\bar X_n)$$
We have $\operatorname{Var}(Z) =2$ and $\operatorname{Var}(\bar X_n) = \frac {\sigma^2}{n}$, which gets us to
$$\operatorname{Var}(\hat h_n)=2\left(\frac {\sigma^2}{n}\right)^2 + 4\theta^2\frac {\sigma^2}{n}+ 4\theta\frac {\sigma^2}{n}\operatorname{Cov}(Z,\bar X_n)$$
For the Covariance we have
$$\operatorname{Cov}(Z,\bar X_n) = E(Z\bar X_n) - E(Z)E(\bar X_n) = E\left(\left[ \frac{\left(\bar X_n -\theta \right) \sqrt{n}} {\sigma} \right]^2\bar X_n\right) - 1\cdot \theta$$
$$= \frac {n}{\sigma^2}E\left(\bar X_n^3 -2\theta \bar X_n^2 + \theta^2\bar X_n\right) -\theta$$
We have $E(\bar X_n^2) = \frac {\sigma^2}{n} + \theta^2$, and $E(\bar X_n) = \theta$. Also, one can verify ($\bar X_n$ follows a normal distribution with non-zero mean) that $$E(\bar X_n^3) = \theta\left(\theta^2+3\frac {\sigma^2}{n}\right) = \theta^3+3\theta\frac {\sigma^2}{n}$$
So
$$ \operatorname{Cov}(Z,\bar X_n) = \frac {n}{\sigma^2}\left(\theta^3+3\theta\frac {\sigma^2}{n} -2\theta \left(\frac {\sigma^2}{n} + \theta^2\right) + \theta^2\theta\right) -\theta =0$$
(what is the intuition behind this result that says that $Z$ and $\bar X_n$ are uncorrelated?)
So the the variance of the estimator (for "large" but finite $n$) is just
$$\operatorname{Var}(\hat h_n)=2\left(\frac {\sigma^2}{n}\right)^2 + 4\theta^2\frac {\sigma^2}{n}$$
Note: One could more quickly consult the tables for the raw moments of a normal distribution with non-zero mean, and derive the variance of the estimator directly from
$$\operatorname{Var}(\hat h_n) = E\left(\bar X_n^4\right) - \left(E(\bar X_n^2)\right)^2$$
...but one would then lose the conclusion that $Z$ and $\bar X_n$ are uncorrelated.
