= Solution
For odd $p$, write $e_p(t)=e^{2\pi it/p}$ and use the <discrete paraboloid>
$$
P=\{(u,v,u^2+v^2):u,v\in\mathbb F_p\},\qquad \sigma=p^{-2}1_P.
$$
Here surface integration means $p^{-2}\sum_P$, whereas ambient <norms> below use unnormalized counting measure on $\mathbb F_p^3$. With the negative-phase transform convention,
$$
\widehat\sigma(x)=p^{-2}\sum_{u,v}e_p(-x_1u-x_2v-x_3(u^2+v^2)).
$$
When $x_3=0$, additive-<orthogonality of characters> makes this one at $x=0$ and zero at all other points of that plane. When $x_3\ne0$, completing the two squares gives
$$
\widehat\sigma(x)=p^{-2}e_p\left(\frac{x_1^2+x_2^2}{4x_3}\right)G(-x_3)^2,\qquad G(t)=\sum_u e_p(tu^2).
$$
The <Quadratic Gauss sum> identity is $G(t)^2=\chi(-1)p$, where $\chi$ is the <Legendre symbol>. For completeness, counting the preimages of a square gives $G(t)=\chi(t)G(1)$ for nonzero $t$. Also $|G(1)|^2=\sum_{u,v}e_p(u^2-v^2)=p$, by the invertible change $(u,v)\mapsto(u-v,u+v)$ and <orthogonality of characters>. Since $\overline{G(1)}=G(-1)=\chi(-1)G(1)$, the asserted square identity follows. Thus for $p\equiv3\pmod4$,
$$
\boxed{\widehat\sigma(x)=\begin{cases}1,&x=0,\\0,&x_3=0,\ x\ne0,\\-p^{-1}e_p((x_1^2+x_2^2)/(4x_3)),&x_3\ne0.\end{cases}}
$$
Using positive phase instead changes the sign inside the character, not the magnitude or any <norm> estimate.
Define $Eg(x)=p^{-2}\sum_{\xi\in P}g(\xi)e_p(x\cdot\xi)$, and define $R^*(2\to q)$ as the smallest constant for
$$
\|Eg\|_{\ell^q(\mathbb F_p^3)}\leq R^*(2\to q)\left(p^{-2}\sum_{\xi\in P}|g(\xi)|^2\right)^{1/2}.
$$
The adjoint relative to these measures is $E^*h(\xi)=\sum_xh(x)e_p(-x\cdot\xi)$. Consequently $EE^*h=h*\widehat\sigma(-\cdot)$, where convolution is the unnormalized ambient sum. Put $K=\widehat\sigma(-\cdot)-1_{\{0\}}$. Its supremum is at most $p^{-1}$, so convolution by $K$ has $\ell^1\to\ell^\infty$ <norm> at most $p^{-1}$.
Under the unnormalized ambient <Fourier transform on a finite group>, the convolution multiplier of $\widehat\sigma(-\cdot)$ is $p^3\sigma=p1_P$, up to the harmless reflection convention. Thus the multiplier of $K$ is $p1_P-1$, with maximum absolute value at most $p$. The <Plancherel theorem> gives $\ell^2\to\ell^2$ <norm> at most $p$. The <Riesz-Thorin interpolation theorem> between these two bounds gives $\ell^{4/3}\to\ell^4$ <norm> at most one for convolution by $K$. Convolution by $1_{\{0\}}$ is the identity, whose $\ell^{4/3}\to\ell^4$ <norm> is at most one for counting measure. Hence $\|EE^*h\|_4\leq2\|h\|_{4/3}$.
By <Lp duality> and <Hölder's inequality>, $\|E^*h\|_{L^2(\sigma)}^2=\langle EE^*h,h\rangle\leq2\|h\|_{4/3}^2$. Taking adjoints proves the <finite-field paraboloid fourth-moment extension estimate>
$$
\boxed{R^*(2\to4)\leq\sqrt2<10.}
$$
In fact this argument works for both residue classes of odd primes.
If $p\equiv1\pmod4$, choose $i\in\mathbb F_p$ with $i^2=-1$. The complete line $\ell=\{(t,it,0):t\in\mathbb F_p\}$ lies in $P$. For $g=1_\ell$ the normalized surface $L^2$ <norm> is $p^{-1/2}$, and <orthogonality of characters> gives $Eg(x)=p^{-1}1_{\{x_1+ix_2=0\}}$. That annihilator plane has $p^2$ points, so the <isotropic-line obstruction to finite-field restriction> gives
$$
\boxed{R^*(2\to q)\geq\frac{p^{2/q-1}}{p^{-1/2}}=p^{2/q-1/2}.}
$$
This tends to infinity along these primes for every $q<4$, as required.
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