Given that

$x\varphi \left(x\right)={\int }_5^x\left(3t^2-2{\varphi }^{\text{'}}\left(t\right)\right)dt$

$\varphi \left(0\right)=4, x>-2$

Differentiating on both the sides and using Newton Leibnitz rule 

$\Rightarrow x{\varphi }^{\text{'}}\left(x\right)+1\varphi \left(x\right)=3x^2-2{\varphi }^{\text{'}}\left(x\right)$

$\because \frac{d}{dx}\left(UV\right) = U\frac{dV}{dx} + V\frac{dU}{dx} ; \frac{d}{dx}{\int }_{f\left(x\right)}^{g\left(x\right)}h\left(x\right)dx = g^{\text{'}}\left(x\right)h\left(g\left(x\right)\right)-f^{\text{'}}\left(x\right)h\left(f\left(x\right)\right)$

$\Rightarrow (x+2){\varphi }^{\text{'}}\left(x\right)+\varphi \left(x\right)=3x^2$

It is a linear differential equation in the form of $\frac{dy}{dx}+Py = Q$

$\Rightarrow I.F.=e^{\int \left(\frac{1}{x+2}\right)dx}=x+2$

$\Rightarrow \varphi \left(x\right)·\left(x+2\right)=\int \left(x+2\right)\left(\frac{3x^2}{x+2}\right)dx$

$\Rightarrow \varphi \left(x\right)·(x+2)=x^3+c$

Given that $\varphi \left(0\right)=4$

$\Rightarrow c=8$

$\Rightarrow \varphi \left(2\right)=\frac{8+8}{4}=4$