The average squared determinant

Idea. Expand the square of the determinant as a double sum over permutations. Because the entries are independent with mean 0, only the terms where both permutations are the same survive, and each of those has expectation 1.

Step 1: expand. $$\det A=\sum _{\sigma }\mathrm{sgn}(\sigma )\prod _{i=1}^3{a}_{i\sigma (i)},$$ so $$\det (A{)}^2=\sum _{\sigma ,\tau }\mathrm{sgn}(\sigma )\mathrm{sgn}(\tau )\prod _{i=1}^3{a}_{i\sigma (i)}{a}_{i\tau (i)}.$$

Step 2: which terms survive. If $\sigma \ne \tau$, there is a row $i$ with $\sigma (i)\ne \tau (i)$. The entry $a_{i\sigma (i)}$ then appears exactly once in the product: no other factor uses it, since in row $i$ the only other entry is $a_{i\tau (i)}$. Independence and mean 0 make the expectation of that term 0. If $\sigma =\tau$, the term is ${\prod }_i{a}_{i\sigma (i)}^2=1$.

Step 3. There are $3!=6$ permutations, so $$\mathbb{E}\left[\det (A{)}^2\right]=6.$$

In general. For an $n\times n$ matrix with independent entries of mean 0 and variance 1, $\mathbb{E}[\det (A{)}^2]=n!$. Since $\mathbb{E}[\det A]=0$, this is also the variance of the determinant.