Chọn A.
Ta có $V=\pi \underset{1}{\overset{\text{e}}{\int }}{\left(x\ln x\right)}^2.\text{d}x=\pi \underset{1}{\overset{\text{e}}{\int }}{x}^2{\ln }^{2}x.\text{d}x$.
Đặt $\left\{\begin{array}{l}u={\ln }^{2}x\\ \text{d}v=x^2\text{d}x\end{array}\right.\Rightarrow \left\{\begin{array}{l}\text{d}u=2.\frac{1}{x}.\ln x.\text{d}x\\ v=\frac{x^3}{3}\end{array}\right.$
Khi đó $V=\pi \left[\frac{x^3}{3}.{\ln }^{2}x\left|{}_{\begin{array}{l}\\ 1\end{array}}^{\begin{array}{l}\text{e}\\ \end{array}}\right.-\underset{1}{\overset{\text{e}}{\int }}\frac{x^3}{3}\frac{2}{x}.\ln x.\text{d}x\right]=\pi \left[\frac{{\text{e}}^3}{3}-\frac{2}{3}\underset{1}{\overset{\text{e}}{\int }}{x}^2\text{.ln }x.\text{d}x\right]$
Đặt $\left\{\begin{array}{l}u=\ln x\\ \text{d}v=x^2\text{d}x\end{array}\right.\Rightarrow \left\{\begin{array}{l}\text{d}u=\frac{1}{x}.\text{d}x\\ v=\frac{x^3}{3}\end{array}\right.$
$\underset{1}{\overset{\text{e}}{\int }}{x}^2\text{.ln }x.\text{d}x=\frac{x^3}{3}.\ln x\left|{}_{\begin{array}{l}\\ 1\end{array}}^{\begin{array}{l}\text{e}\\ \end{array}}\right.-\underset{1}{\overset{\text{e}}{\int }}\frac{x^3}{3}\text{.}\frac{1}{x}.\text{d}x=\frac{{\text{e}}^3}{3}-\underset{1}{\overset{\text{e}}{\int }}\frac{x^2}{3}\text{.d}x=\frac{{\text{e}}^3}{3}-\frac{x^3}{9}\left|{}_{\begin{array}{l}\\ 1\end{array}}^{\begin{array}{l}\text{e}\\ \end{array}}\right.=\frac{{\text{e}}^3}{3}-\frac{{\text{e}}^3}{9}+\frac{1}{9}=\frac{{\text{2e}}^3+1}{9}$.

Vậy $V=\pi \left[\frac{{\text{e}}^3}{3}-\frac{2}{3}\left(\frac{2{\text{e}}^3+1}{9}\right)\right]=\frac{\pi \left(5{\text{e}}^3-2\right)}{27}$.