We can expand the inner product by using two additional integration variables:
\begin{align}
\langle 2|2\rangle&=\langle 0|\int d^3x_3 d^3x_4 \varphi^*(\mathbf{x_3},\mathbf{x_4})\psi(\mathbf{x_3})\psi(\mathbf{x_4})\int d^3x_1 d^3x_2 \varphi(\mathbf{x_1},\mathbf{x_2})\psi^\dagger(\mathbf{x_1})\psi^\dagger(\mathbf{x_2})|0\rangle\\
&=\int d^3x_1 d^3x_2 d^3x_3 d^3x_4 \varphi^*(\mathbf{x_3},\mathbf{x_4}) \varphi(\mathbf{x_1},\mathbf{x_2})\langle 0|\psi(\mathbf{x_3})\psi(\mathbf{x_4})\psi^\dagger(\mathbf{x_1})\psi^\dagger(\mathbf{x_2})|0\rangle.
\end{align} Using reasoning like in your previous question, we can evaluate the expectation value
\begin{aligned}
\langle 0|\psi(\mathbf{x_3})\psi(\mathbf{x_4})\psi^\dagger(\mathbf{x_1})\psi^\dagger(\mathbf{x_2})|0\rangle&=\langle 1_{\mathbf{x}_3},1_{\mathbf{x}_4}|1_{\mathbf{x}_1},1_{\mathbf{x}_2}\rangle \\
&=\left[\delta({\mathbf{x}_3}-{\mathbf{x}_1})\delta({\mathbf{x}_4}-{\mathbf{x}_2})+
\delta({\mathbf{x}_3}-{\mathbf{x}_2})\delta({\mathbf{x}_4}-{\mathbf{x}_1})\right],
\end{aligned} where we have used the second-quantized notation saying that, for example the state $|1_{\mathbf{x}_1},1_{\mathbf{x}_2}\rangle$ has $1$ boson with coordinates $\bf{x}_1$ and one with coordinates $\bf{x}_2$ (if both have the same coordinates, we get an extra factor of $\sqrt{2}$ on both the bra and the ket, which is accounted for by the sum over the two delta functions).
We can evaluate the integrals over $\mathbf{x}_1$ and $\bf{x}_2$ using the delta functions (I guessed wrong about the initial labels so we have to adjust the labels) to find
\begin{align}
\langle 2|2\rangle
&=\int d^3x_3 d^3x_4 \varphi^*(\mathbf{x_3},\mathbf{x_4}) \left[\varphi(\mathbf{x_3},\mathbf{x_4})+\varphi(\mathbf{x_4},\mathbf{x_3})\right]\\
&=\int d^3x_1 d^3x_2 \varphi^*(\mathbf{x_1},\mathbf{x_2}) \left[\varphi(\mathbf{x_1},\mathbf{x_2})+\varphi(\mathbf{x_2},\mathbf{x_1})\right].
\end{align} This is exactly the desired expression, up to a typo in $\phi^*(\mathbf{x}_1,\mathbf{x}_2)$.
