= Solution
The squared <norm> of $p_\ell(x)e^{-x^2/4}$ in $L^2(\mathbb R,dx)$ is $\sqrt{2\pi}\,h_\ell$, where $(h_0,h_1,h_2)=(1,1,2)$. Choose positive normalizing constants
$$
\boxed{\kappa_0=\kappa_1=(2\pi)^{1/4},\qquad
\kappa_2=\sqrt2(2\pi)^{1/4}.}
$$
Then $\phi_0,\phi_1,\phi_2$ are an <orthonormal set> for <Lebesgue measure>. Expanding the finite sums gives
$$
\begin{aligned}
\int\widetilde K_3(x,y)\widetilde K_3(y,z)\,dy
&=\sum_{\ell,m=0}^2\phi_\ell(x)\phi_m(z)
\int\phi_\ell(y)\phi_m(y)\,dy\\
&=\sum_{\ell=0}^2\phi_\ell(x)\phi_\ell(z)
=\widetilde K_3(x,z).
\end{aligned}
$$
Thus this is an <orthogonal polynomial projection kernel>, explicitly
$$
\boxed{\widetilde K_3(x,y)
=\frac{e^{-(x^2+y^2)/4}}{\sqrt{2\pi}}
\left[1+xy+\frac{(x^2-1)(y^2-1)}2\right].}
$$
The positive constants above are forced by the reproducing identity: before normalization, its convolution multiplies each independent term $\phi_\ell(x)\phi_\ell(z)$ by $\sqrt{2\pi}h_\ell/\kappa_\ell^2$. Each multiplier must equal one. Changing the sign of a real $\kappa_\ell$ leaves the kernel unchanged.
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