Let $\frac{d}{dx}F\left(x\right)=\frac{e^{\sin x}}{x},x>0$, if $\underset{1}{\overset{4}{\int }}\frac{3}{x}\cdot e^{\sin x^3}dx=F\left(k\right)-F\left(1\right)$, then one of the possible values of $k=$

$\frac{d}{dx}F\left(x\right)=\frac{e^{\sin x}}{x}$

$\Rightarrow F\left(x\right)=\int \frac{e^{\sin x}}{x}dx$

$F\left(k\right)-F\left(1\right)=\underset{1}{\overset{4}{\int }}\frac{3}{x}\cdot e^{\sin x^3}\cdot dx$

$=\underset{1}{\overset{64}{\int }}{e}^{\sin t}\cdot \frac{3}{x}\cdot \frac{1}{3x^2}dt$

$=\underset{1}{\overset{64}{\int }}\frac{e^{\sin t}}{t}dt$

Put $x^3=t$

$\Rightarrow 3x^{2}dx=dt$

$\Rightarrow dx=\frac{1}{3x^2}dt$

$x=1\Rightarrow t=1$

$x=4\Rightarrow t=64$

$={F\left(t\right)]}_1^{64}$

$=F\left(64\right)-F\left(1\right)$

$\therefore K=64$.