Given, equation
$$
\begin{aligned}
& 6 x^6-25 x^5+31 x^4-31 x^2+25 x-6=0 \\
& \Rightarrow 6\left(x^6-1\right)-25 x\left(x^4-1\right)+31 x^2\left(x^2-1\right)=0 \\
& \Rightarrow\left(x^2-1\right)\left[6\left(x^4+x^2+1\right)-25 x\left(x^2+1\right)+31 x^2\right]=0 \\
& \Rightarrow\left(x^2-1\right)\left[6 x^4-25 x^3+37 x^2-25 x+6\right]=0
\end{aligned}
$$
Either $x^2-1=0$
$$
\begin{aligned}
& \Rightarrow x=+1,-1 \\
& \text { or } 6 x^4-25 x^3+37 x^2-25 x+6=0 \\
& \Rightarrow 6\left(x^4+1\right)-25 x\left(x^2+1\right)+37 x^2=0 \\
& \Rightarrow 6\left(x^2+\frac{1}{x^2}\right)-25\left(x+\frac{1}{x}\right)+37=0 \\
& \Rightarrow 6\left(x+\frac{1}{x}\right)^2-25\left(x+\frac{1}{x}\right)+25=0 \\
& \Rightarrow 6\left(x+\frac{1}{x}\right)^2-15\left(x+\frac{1}{x}\right)-10\left(x+\frac{1}{x}\right)+25=0 \\
& \Rightarrow 3\left(x+\frac{1}{x}\right)\left[2\left(x+\frac{1}{x}\right)-5\right]-5\left[2\left(x+\frac{1}{x}\right)-5\right]=0 \\
& \left.\Rightarrow 2\left(x+\frac{1}{x}\right)-5\right]\left[3\left(x+\frac{1}{x}\right)-5\right]=0 \\
& \Rightarrow \text { Either } 2\left(x+\frac{1}{x}\right)=5 \text { or } 3\left(x+\frac{1}{x}\right)=5 \\
& \Rightarrow 2 x^2-5 x+2=0 \text { or } 3 x^2-5 x+3=0 \\
& \Rightarrow 2 x^2-4 x-x+2=0 \\
& \because 3 x^2-5 x+3=0 \text { has no solution because } D < 0 . \\
& \Rightarrow 2 x(x-2)-1(x-2)=0 \\
& \Rightarrow x=2, \frac{1}{2}
\end{aligned}
$$

So, sum of all the rational roots
$$
\lambda=1-1+2+\frac{1}{2}=2.5
$$

Hence option (d) is correct.