Given,

$f\left(x\right)=\left\{\begin{array}{c}\frac{1}{x}, x\ge 2\\ a{x}^2+2 b, -2<x<2\\ -\frac{1}{x}, x\le -2\end{array}\right.$

Also given $f\left(x\right)$ is differentiable on $\mathrm{R}$

So, finding LHD and RHD at $x=2$ we get,

$-\frac{1}{x^2}=2ax$

$\Rightarrow \frac{-1}{4}=4\mathrm{a}$

$\Rightarrow \mathrm{a}=\frac{-1}{16}$

Now, for function to be continuous at $x=2$ we get,

$\frac{1}{2}=4a+2b$

$\Rightarrow \frac{1}{2}=\frac{-4}{16}+2b$

$\Rightarrow b=\frac{3}{8}$

Hence, $48\left(a+b\right)=48\left(\frac{3}{8}-\frac{1}{16}\right)$

$\Rightarrow 48\left(a+b\right)=48\left(\frac{5}{16}\right)=15$