The formula for the integration,

$I_m={\int }_0^{\infty }{e}^{-x}{\sin }^{m}xdx,$ where $m>2$

$\Rightarrow I_m=\frac{m(m-1)}{1+m^2}{I}_{m-2}$

Now,

$I_6={\int }_0^{\infty }{e}^{-x}\sin 6xdx$

$\Rightarrow I_6=\frac{6\left(5\right)}{1+6^2}{I}_4$

$\Rightarrow I_6=\frac{30}{37}{I}_4$

Similarly,

$I_4={\int }_0^{\infty }{e}^{-x}\sin 4xdx$

$\Rightarrow I_4=\frac{4\left(3\right)}{1+4^2}{I}_2$

$\Rightarrow I_4=\frac{12}{17}{I}_2$

Similarly,

$I_6={\int }_0^{\infty }{e}^{-x}\sin 6xdx$

$\Rightarrow I_2=\frac{2\left(1\right)}{1+2^2}{I}_0$

$\Rightarrow I_2=\frac{2}{5}{I}_0$

And

$I_0={\int }_0^{\infty }{e}^{-x}{\sin }^{0}xdx$

$\Rightarrow I_0={\left[-e^{-x}\right]}_0^{\infty }=1$

The value of $I_6$ is

$I_6=\frac{30}{37}\times \frac{12}{17}\times \frac{2}{5}\times 1=\frac{144}{629}$.