If $a{x}^2+\frac{b}{x}\ge c$ for all positive $x$, where $a,b>0$, then

Given function is $a{x}^2+\frac{b}{x}\ge c$

Assume the given function as  $y=a{x}^2+\frac{b}{x}\ge c$

Differentiate w.r.t x

$$\begin{gathered} \frac{dy}{dx}=2ax-\frac{b}{x^2} \\ \frac{dy}{dx}=\frac{2a{x}^3-b}{x^2} \end{gathered}$$

Differentiate again w.r.t $x$

$\frac{d^{2}y}{d{x}^2}=2a+\frac{2b}{x^3} >0$

Equate $\frac{dy}{dx}=0$

$x={\left(\frac{b}{2a}\right)}^{\frac{1}{3}}$

Observe that $\frac{d^{2}y}{d{x}^2}\left(x={\left(\frac{b}{2a}\right)}^{\frac{1}{3}}\right)$ is positive.

$f\left(x\right)$ has minimum value at $x={\left(\frac{b}{2a}\right)}^{\frac{1}{3}}$

Substitute the value of $x$ in the given  function.

$$\begin{gathered} a{x}^2+\frac{b}{x}\ge c \\ a{\left(\frac{b}{2a}\right)}^{\frac{2}{3}}+\frac{b}{{\left({\displaystyle \frac{b}{2a}}\right)}^{{\displaystyle \frac{1}{3}}}}\ge c \\ a\left(\frac{b}{2a}\right)+b\ge c{\left(\frac{b}{2a}\right)}^{\frac{1}{3}} \\ \frac{3b}{2}\ge c{\left(\frac{b}{2a}\right)}^{\frac{1}{3}} \end{gathered}$$

Cube on both sides,

$$\begin{gathered} \frac{27b^3}{8}\ge c^3\left(\frac{b}{2a}\right) \\ \frac{27b^3}{8}\times \frac{2a}{b}\ge c^3 \\ 27a{b}^2\ge 4c^3 \end{gathered}$$

Therefore, the solution of the given function is $27b^2\ge 4c^3$.

Hence, the correct answer is option $\left(1\right)$.