Given $\underset{x\to 0}{\lim }\frac{\cos \left(\sin x\right)-\cos x}{x^4}=f\left(0\right)$

$\Rightarrow \underset{x\to 0}{\lim }\frac{2\sin \left(\frac{\sin x+x}{2}\right)\sin \left(\frac{x-\sin x}{2}\right)}{x^4}=\frac{1}{k}$

$\Rightarrow \underset{x\to 0}{\lim }\frac{2\sin \left(\frac{\sin x+x}{2}\right)\sin \left(\frac{x-\sin x}{2}\right)}{\left({\displaystyle \frac{\sin x+x}{2}}\right)\left({\displaystyle \frac{x-\sin x}{2}}\right)x^4}\times \left(\frac{\sin x+x}{2}\right)\left(\frac{x-\sin x}{2}\right)=\frac{1}{k}$

$\Rightarrow \underset{x\to 0}{\lim }2\left(\frac{\sin x+x}{2x}\right)\left(\frac{x-\sin x}{2x^3}\right)=\frac{1}{k}, \left(\because \underset{x\to 0}{\lim }\frac{\sin x}{x}=1\right)$

$\Rightarrow 2\left(\underset{x\to 0}{\lim }\left(\frac{\sin x+x}{2x}\right)\right)\left(\underset{x\to 0}{\lim }\left(\frac{x-\sin x}{2x^3}\right)\right)=\frac{1}{k}$

Apply L'Hospital rule for an indeterminate form of $\frac{0}{0}$, we get, 

$\Rightarrow 2\times \frac{(1+1)}{2}\times \frac{1}{2}\times \frac{1}{6}=\frac{1}{ k}$

$\Rightarrow k=6$