Random variable $Y$ has mean $0$ and variance $\frac{1}{w}$.
I am trying to find the distribution of the random variable $wY^2$.
For the mean,
$E(Y^2)=(E(Y))^{2}+Var(Y)=0+\frac{1}{w}=\frac{1}{w}$.
However, I am struggling with the $var(Y^{2})$ because it would seem that I would need to know $E(Y^4)$, which I in turn cannot calculate without $var(Y^{2})$.
On
Assume $Y$ is a normal random variable. Its 4th moment is evaluated with
\begin{equation} \mathbb{E}[Y^4] = \int_{-\infty}^\infty Y^4 f(Y) \, \mathrm{d}x = \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^\infty Y^4 \mathrm{e}^{-\frac{Y^2}{2\sigma^2}} \, \mathrm{d}Y. \end{equation}
Use integration by parts to get,
$$ \mathbb{E}[Y^4] =\frac{3\sigma^2}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^\infty Y^2 \mathrm{e}^{-\frac{Y^2}{2\sigma^2}} \, \mathrm{d}Y =3\sigma^4 = \frac{3}{w^2}$$
where $\sigma^2=1/w$ is used.
Unfortunately, I don't think that there is much more you can do than write $$V(X^2) = \mathbb{E}(X^4) - \sigma^4 - \mu^4 - 2\sigma^2\mu^2 $$ in general, if you don't know the underlying distribution.
In your case this reduces to $$V(X^2) = \mathbb{E}(X^4) - \frac{1}{w^4}. $$