let $X \sim N(0,\sigma^2)$

now let $Y=X^2$

what is $E(Y)$ and $V(Y)$

so $E(Y)=E(X^2)=\sigma^2$

but I can't find $V(Y)$

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    $\begingroup$ Surely you can find $V(Y)$ from the definition, because the moments of $Y$ bear a simple relationship to those of $X:$ $$E[Y^k] = E[(X^2)^k] = E[X^{2k}]$$ shows the raw moments of $Y$ are the corresponding raw even moments of $X.$ $\endgroup$ – whuber Mar 13 at 13:47

Using the definition of variance, you have: $$\operatorname{var}(Y)=E[X^4]-E[X^2]^2$$ You need the 4-th non-central moment of normal RV. Since, the mean is $0$, you can also use central moment, which is $3\sigma^4$. Subtracting $E[X^2]^2=\sigma^4$ makes variance of $Y$ $2\sigma^4$.

Another way is to use the relation between normal, chi-squared and gamma distributions. The squared standard normal is chi-squared distribution with $k=1$, which is at the same time Gamma RV. So, let $X=\sigma Z$, then $Z^2\sim\chi^2(1)\rightarrow \sigma^2 Z^2=X^2 \sim \Gamma(k=1/2,\theta=2\sigma^2)$. So, $Y$ is actually Gamma distributed. Gamma variance is $\operatorname{var}(Y)=k\theta^2=2\sigma^4$.

Another way can be simply evaluating the integral (let $c=\frac{1}{\sqrt{2\pi}\sigma}$):

$$\begin{align}E[X^4]&=c\int_{-\infty}^\infty x^4 \exp(-x^2/2\sigma^2)dx\\&=c\underbrace{\int_{-\infty}^\infty x^4 \exp(-x^2/2\sigma^2)dx}_{u=x^3,dv=x\exp(-x^22\sigma^2)}\\&=3\sigma^2\underbrace{\int_{-\infty}^{\infty}cx^2\exp(-x^2/2\sigma^2)dx}_{E[X^2]}\\&=3\sigma^4\end{align}$$

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