= Solution
At the <standard normal distribution>, symmetry and the odd <Huber score> give population root zero. For $0<b<\infty$, the derivative of the clipped score is $\mathbf1_{\{|x|<b\}}$ except at its two corners, which have zero normal probability. Put $m_b=2\Phi(b)-1>0$ and write $\phi$ for the <standard normal density>. The <sandwich variance of an M-estimator> has numerator
$$
\mathbb E\psi_b(Z)^2=\int_{-b}^b z^2\phi(z)dz+2b^2[1-\Phi(b)].
$$
Using $\phi'(z)=-z\phi(z)$ and integration by parts,
$$
\int_{-b}^b z^2\phi(z)dz=-2b\phi(b)+2\Phi(b)-1.
$$
Therefore
$$
\boxed{V(\psi_b,\Phi)=\frac{2\Phi(b)-1-2b\phi(b)+2b^2[1-\Phi(b)]}{[2\Phi(b)-1]^2}.}
$$
As checks, it tends to $1$ as $b\to\infty$, the <sample mean> limit, and to $\pi/2$ as $b\downarrow0$, the <sample median> limiting <variance>. The estimator <variance> for a sample of size $n$ is approximately this quantity divided by $n$.
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