1. We need to evaluate $I=\int_{1/3}^{3}|\ln x|\,dx.$

2. Since

• $\ln x<0$ for $0<x<1$,
• $\ln x>0$ for $x>1$, we split the integral at $x=1$: $I=\int_{1/3}^{1}(-\ln x)\,dx+\int_{1}^{3}\ln x\,dx.$

3. Use the standard result $\int \ln x\,dx=x\ln x-x+C.$ Hence, $\int (-\ln x)\,dx=-(x\ln x-x)=-x\ln x+x.$

4. First part: $I_1=\int_{1/3}^{1}(-\ln x)\,dx=[-x\ln x+x]_{1/3}^{1}.$ Now,

• at $x=1$: $-1\cdot \ln 1+1=1$,
• at $x=1/3$: $-\frac13\ln\left(\frac13\right)+\frac13=\frac13\ln 3+\frac13.$ So, $I_1=1-\left(\frac13\ln 3+\frac13\right)=\frac23-\frac13\ln 3.$

5. Second part: $I_2=\int_1^3 \ln x\,dx=[x\ln x-x]_1^3.$ Now,

• at $x=3$: $3\ln 3-3$,
• at $x=1$: $0-1=-1$. Thus, $I_2=(3\ln 3-3)-(-1)=3\ln 3-2.$

6. Add both parts: $I=I_1+I_2=\left(\frac23-\frac13\ln 3\right)+(3\ln 3-2).$ So, I=-\frac43+\frac83\ln 3=rac83\ln 3-\frac43. Factor $\frac43$: $I=\frac43(2\ln 3-1).$

7. We are given $I=\frac{m}{n}\ln\left(\frac{n^2}{e}\right).$ Now, $\ln\left(\frac{n^2}{e}\right)=\ln(n^2)-1=2\ln n-1.$ We want $\frac{m}{n}(2\ln n-1)=\frac43(2\ln 3-1).$ This matches directly if $n=3,\qquad \frac{m}{n}=\frac43 \Rightarrow m=4.$ Clearly, $m=4$ and $n=3$ are coprime.

8. Therefore, ${m^2+n^2-5}=4^2+3^2-5=16+9-5=20.$

So the required integer is $\boxed{20}.$