(81)

Given, $\underset{x\to 0}{\lim }\frac{a{x}^2{e}^x-b{\log }_e(1+x)+cx{e}^{-x}}{x^2\sin x}=1$

$\Rightarrow \underset{x\to 0}{\lim }\frac{a{x}^2(1+\frac{x}{1!}+\frac{x^2}{2!}+\dots )-b(x-\frac{x^2}{2}+\frac{x^3}{3}\dots )+cx(1-\frac{x}{1!}+\frac{x^2}{2!}\dots )}{x^2(x-\frac{x^3}{3!}+\frac{x^5}{5!}\dots )}=1$

$\Rightarrow \underset{x\to 0}{\lim }(a{x}^2+a{x}^3+\frac{{\displaystyle a{x}^4}}{{\displaystyle 2!}}+\dots )-bx+\frac{b{x}^2}{2}-\frac{b{x}^3}{3}\dots +cx-c{x}^2+\frac{c{x}^3}{2!}\dots =\underset{x\to 0}{\lim }{x}^3-\frac{{\displaystyle x^5}}{{\displaystyle 3!}}+\frac{{\displaystyle x^7}}{{\displaystyle 5!}}\dots$

Comparing the coefficient of $x^3,$ we get

$a-\frac{b}{3}+\frac{c}{2}=1$                                                          ...(i)

Comparing the coefficient of $x^2,$ we get

$a+\frac{b}{2}-c=0$                                                            ...(ii)

Comparing the coefficient of $x,$ we get 

$-b+c=0$                                                                 ...(iii)

On solving (i), (ii) and (iii), we get

$a=\frac{3}{4}, b=c=\frac{3}{2}$

$\therefore 16(a^2+b^2+c^2)=16[\frac{{\displaystyle 9}}{{\displaystyle 16}}+\frac{{\displaystyle 9}}{{\displaystyle 4}}+\frac{{\displaystyle 9}}{{\displaystyle 4}}]=16\left[\frac{{\displaystyle 9+36+36}}{{\displaystyle 16}}\right]=81$