Let, $f\left(x\right)=a{x}^3+b{x}^2+cx+d$

Diff w. r. t.  '$x$'

$f^{\text{'}}\left(x\right)=3a{x}^2+2bx+c$

Again diff. w. r. t.  '$x$'

$f^{"}\left(x\right)=6ax+2b$

$f^{"}\left(-1\right)=0$

$-6a+2b=0$

$b=3a$

$f\text{'}\left(1\right)=0$

$3a+6a+c=0$

$c=-9a$

$f\left(1\right)=-10$ 

$-5a+d=-10 \dots \left(\mathrm{i}\right)$

$f(-1)=6$ 

$11a+d=6 \dots \left(\mathrm{ii}\right)$

From equations $\left(\mathrm{i}\right)-\left(\mathrm{ii}\right)$

We get $a=1, d=-5, b=3, c=-9$

Then, $f\left(x\right)=x^3+3x^2-9x-5$

Hence,  $f\left(3\right)=27+27-27-5$

$f\left(3\right)=22$