(c)

We have, $3 x^3-25 x+n=0$

Let $f(x)=3 x^3-25 x+n$

$f^{\prime}(x) =9 x^2-25$

$\text { Put } f^{\prime}(x) =0$

$9 x^2-25 =0$

$x =\pm \frac{5}{3} \Rightarrow x_1=\frac{-5}{3}, x_2=\frac{5}{3}$

$f(x)$ has three real roots.

$\therefore\left(x_1\right) f\left(x_2\right) < 0$

$\therefore\left(3\left(\frac{-5}{3}\right)^3-25 \left(\frac{-5}{3}\right)+n\right)$

$\left(3\left(\frac{5}{3}\right)^3-25\left(\frac{5}{3}\right)+n\right) < 0$

$\left(\frac{-125}{9}+\frac{125}{3}+n\right)\left(\frac{125}{9}-\frac{125}{3}+n\right) < 0$

$\left(n+\frac{250}{9}\right)\left(n-\frac{250}{9}\right) < 0$

$n \in\left(\frac{-250}{9}, \frac{250}{9}\right]$

Hence, $n \in I$

$\therefore$ Total integer is $55$