The correct option is $\frac{4e^2}{3}$.

Step 1: Understand the given integral equation

We are given the following equation for $x\in [1,\infty )$:

$3{\int }_1^{x}f\left(t\right)dt=xf\left(x\right)-\frac{x^3}{3}$

We are also given the initial condition $f\left(1\right)=\frac{1}{3}$.

Step 2: Differentiate both sides with respect to x

To eliminate the integral, we differentiate both sides of the equation with respect to $x$ using the Fundamental Theorem of Calculus (Leibniz Rule) on the left side and the product rule on the right side:

$\frac{d}{dx}\left(3{\int }_1^{x}f\right(t\left)dt\right)=\frac{d}{dx}\left(xf\right(x)-\frac{x^3}{3})$

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

Rearranging the terms, we get:

$2f\left(x\right)=x{f}^{\prime }\left(x\right)-x^2$

$x{f}^{\prime }\left(x\right)-2f\left(x\right)=x^2$

Dividing both sides by $x$ (since $x\ge <!--\; U+2265\; -->1$):

$f^{\prime }\left(x\right)-\frac{2}{x}f\left(x\right)=x$

Step 3: Solve the Linear Differential Equation

This is a first-order linear differential equation of the form $f^{\prime }\left(x\right)+P\left(x\right)f\left(x\right)=Q\left(x\right)$, where:

$P\left(x\right)=-\frac{2}{x}$ and $Q\left(x\right)=x$

First, find the Integrating Factor (I.F.):

$I.F.=e^{\int <!--\; U+222B\; -->-\frac{2}{x}dx}=e^{-2{log}_{e}x}=e^{{log}_e\left(x^{-2}\right)}=\frac{1}{x^2}$

Multiplying the differential equation by the Integrating Factor gives:

$\frac{d}{dx}\left[f\right(x)·<!--\; U+22C5\; -->\frac{1}{x^2}]=x·<!--\; U+22C5\; -->\frac{1}{x^2}=\frac{1}{x}$

Integrating both sides with respect to $x$:

$\frac{f\left(x\right)}{x^2}={log}_{e}x+C$

$f\left(x\right)=x^2({log}_{e}x+C)$

Step 4: Find the constant of integration C

Substitute the given condition $f\left(1\right)=\frac{1}{3}$ into the function:

$f\left(1\right)=1^2({log}_{e}1+C)=\frac{1}{3}$

$0+C=\frac{1}{3}\Rightarrow <!--\; U+21D2\; -->C=\frac{1}{3}$

Thus, the function is:

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

Step 5: Calculate f(e)

Substitute $x=e$ into the function:

$f\left(e\right)=e^2({log}_{e}e+\frac{1}{3})$

Since ${log}_{e}e=1$:

$f\left(e\right)=e^2(1+\frac{1}{3})=e^2\left(\frac{4}{3}\right)=\frac{4e^2}{3}$