$f(x)=a_{0}+a_{1}x+a_{2}x^{2}+\dots +a_{n}x^{n}$
Hence $f(A)=a_{0}+a_{1}A+a_{2}A^{2}+ \dots +a_{n}A^{n}$
Characteristic equation $f(\lambda)=a_{0}+a_{1}\lambda+a_{2}\lambda^{2}+ \dots+a_{n}\lambda^{n}$
Lets take $ n =2$
$f(\lambda)=a_{0}+a_{1}\lambda+a_{2}\lambda^{2}$
$f(\lambda)= a_{2}\lambda^{2} + a_{1}\lambda + a_{0}$
Lets say roots are $\lambda_{1},\lambda_{2}$
$\mid A \mid = \lambda_{1}\times \lambda_{2} = \dfrac{a_{0}}{a_{2}}$
Lets take $ n =3$
$f(\lambda)=a_{0}+a_{1}\lambda+a_{2}\lambda^{2} + a_{3}\lambda^{3}$
$f(\lambda)= a_{3}\lambda^{3} + a_{2}\lambda^{2} + a_{1}\lambda + a_{0}$
Lets say roots are $\lambda_{1},\lambda_{2},\lambda_{1}$
$\mid A \mid = \lambda_{1}\times \lambda_{2}\times \lambda_{3} = -\:\dfrac{a_{0}}{a_{3}}$
$\text{det(A) = Product of eigen values} = \lambda_{1}\times \lambda_{2}\times \dots \times \lambda_{n}$
Hence $det(A)= (-1)^{n}\dfrac{a_{0}}{a_{n}}\:\: \text{(Product of roots of equation)}$
So, the correct answer is $(C).$