Suppose $Z , \epsilon \sim N(0, 1)$ are independent Gaussian random variables. Let $a \ll 1$ be a small positive number. Let $W = aZ + \epsilon$. It can be show that \begin{align} \mathbb{E} [ W^2 (Z^2 - 1)] = 2 a^2. \end{align} Now suppose I truncate random variable $W$. I was wondering what truncation level $R$ should be such that \begin{align} \mathbb{E} [ W^2 \cdot \mathbf{1}\{ | W| > R\} \cdot (Z^2 - 1)] \leq a^2. \end{align} Can we set $R$ to be a constant? This seems an easy problem, but computing the expectation seems not trivial.

There is a simple but often efficient trick to facilitate such computations. Let $H_k$ be the normalized Hermite polynomials and let $\Phi,\Psi$ be any two functions square integrable with respect to the Gaussian measure $\gamma$. Write $\Phi=\sum_k\varphi_kH_k$, $\Psi=\sum_k\psi_kH_k$. Then, if $Y,Z$ are jointly Gaussian with variance $1$ and covariance $\alpha$, we have $$ E(\Phi(Y)\Psi(Z))=\sum_k \varphi_k\psi_k\alpha^k\,. $$ In your case $\Psi$ is essentially $H_2$ (up to normalization) and $\Phi_r(x)=x^2\chi_{[-r,r]^c}(x)$, so putting $Y=(a^2+1)^{-1/2}W$, we see that we just need to drop the integral $2\varphi_2(r)=\int_{\mathbb R} \Phi_r(x)(x^2-1)\,d\gamma(x)$ from its initial value $2$ at $r=0$ to $1$. That corresponds to some fixed level $r_0>0$ and the answer to the original problem is $R=r_0\sqrt{a^2+1}$.