$\dfrac{\ln x}{x}$

comparing $a,b$ is the same as comparing $6a,6b$ , that is comparing $3\ln(2),2\ln(3)= \ln(8),\ln(9)$ . the latter is obviously bigger meaning $b>a$

comparing $a,c$ is the same as comparing $10a,10b$ , that is comparing $5\ln(2),2\ln(5) = \ln(32),\ln(25)$ . the former is obviously bigger meaning $a>c$

compounding these we get $\boxed{b>a>c}$

Consider $f(x) = \dfrac {\ln x}x$ . Then $f'(x) = \dfrac d{dx} f(x) = \dfrac {1-\ln x}{x^2}$ , $\implies f'(e) = 0$ . And $f''(x) = \dfrac {2\ln x - 3}{x^3}$ , then $f''(e) < 0$ . This implies that $f(x)$ is maximum when $x=e \approx 3$ . This means that $f(x)$ decreases continuously from the maximum $f(e)$ to $f(3)$ , $f(4)$ , $f(5) \cdots f(\infty)$ . That is:

$\begin{aligned} f(3) > & f(4) > f(5) \\ \frac {\ln 3}3 > & {\color{#3D99F6}\frac {\ln 4}4} > \frac {\ln 5}5 \\ \frac {\ln 3}3 > & {\color{#3D99F6}\frac {\ln 2}2} > \frac {\ln 5}5 \\ \implies b > & \ \ a \ \ > c \end{aligned}$