Given,

$x^{2}f\left(x\right)-x=4{\int }_0^{x}t f\left(t\right) dt$

Now Differentiating both side w.r.t. $x$ we get,

$x^2{f}^{\text{'}}\left(x\right)+2xf\left(x\right)-1=4x f\left(x\right)$

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

$\Rightarrow \frac{dy}{dx}-\frac{2}{x}y=\frac{1}{x^2}$ $\left(\mathrm{Let} y=f\left(x\right)\frac{dy}{dx}=f^{\text{'}}\left(x\right)\right)$

Which is a linear differential equation,

So, $\mathrm{I}.\mathrm{F} =e^{\int -\frac{2}{\mathrm{x}}dx}=e^{-2\ln x}=\frac{1}{x^2}$

Now solution of the differential equation is given by,

$\frac{y}{x^2}=\int \frac{1}{x^4}dx+C$

$\Rightarrow \frac{y}{x^2}=-\frac{1}{3x^3}+C$

Now given, $f\left(1\right)=\frac{2}{3}$

So, $\frac{2}{3}=-\frac{1}{3}+C$

$\Rightarrow C=1$

Hence, the function will be,

$f\left(x\right)=-\frac{1}{3x}+x^2$

So, the required value is given by,

$18 f\left(3\right)=18\left[-\frac{1}{9}+9\right]=160$