Given:- $f\left(x\right)=a{x}^3+b{x}^2-\frac{18}{5}x+\frac{19}{10} ....\left(1\right)$

$\therefore f^{\text{'}}\left(x\right)=3a{x}^2+2bx-\frac{18}{5} ...\left(2\right)$

Also, given that $f\left(x\right)$ has maximum value $10$ at $x=-3$

$\therefore f^{\text{'}}\left(-3\right)=3a{\left(-3\right)}^2+2b\left(-3\right)-\frac{18}{5}=0$

$\Rightarrow 27a-6b=\frac{18}{5} . ...\left(3\right)$

Also, $f\left(x\right)$ has minimum value at $x=2$

$\therefore f^{\text{'}}\left(2\right)=0$

$\Rightarrow 3a{\left(2\right)}^2+2b\left(2\right)-\frac{18}{5}=0$

$\Rightarrow 12a+4b=\frac{18}{5}... \left(4\right)$

equation $\left(3\right)\times 2 +$equation $\left(4\right)\times 3,$we get

$54a+36a=\frac{36}{5}+\frac{54}{5}$

$\therefore a=\frac{1}{5}$

Substituting $a=\frac{1}{5}$in $\left(4\right)$, we get

$\frac{12}{5}+4b=\frac{18}{5}$

$\therefore 4b=\frac{6}{5}$

$\therefore b=\frac{3}{10}$

$\therefore f\left(x\right)$becomes,

$\therefore f\left(x\right)=\frac{1}{5}{x}^3+\frac{3}{10}{x}^2-\frac{18}{5}x+\frac{19}{10}$

Now,  $f\left(1\right)=\frac{1}{5}+\frac{3}{10}-\frac{18}{5}+\frac{19}{10}$

$\therefore f\left(1\right)=\frac{-6}{5}$

Hence, option $\left(2\right)$ is correct.