1. We need

Since $x^2\sin x \sim x^3 \quad (x\to 0),$ the numerator must also have leading term exactly $x^3$ with coefficient $1$.

2. Expand each term near $x=0$:

• For $e^x$, $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots$ So $a x^2 e^x=a x^2+a x^3+\frac{a}{2}x^4+\cdots$

• For $\ln(1+x)$, $\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$ So $-b\ln(1+x)=-bx+\frac{b}{2}x^2-\frac{b}{3}x^3+\frac{b}{4}x^4+\cdots$

• For $e^{-x}$, $e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+\cdots$ So $cxe^{-x}=cx-cx^2+\frac{c}{2}x^3-\frac{c}{6}x^4+\cdots$

3. Add the numerator terms:

Thus,

4. Since denominator is of order $x^3$, for the limit to be finite we must have coefficients of $x$ and $x^2$ equal to $0$:

$-b+c=0 \implies c=b$ $a+\frac b2-c=0$ Using $c=b$, $a+\frac b2-b=0 \implies a=\frac b2.$

5. Now the coefficient of $x^3$ in numerator must be $1$, because $x^2\sin x = x^3+O(x^5).$ So, $a-\frac b3+\frac c2=1.$ Substitute $c=b$ and $a=\frac b2$: $\frac b2-\frac b3+\frac b2=1$ $b-\frac b3=1$ $\frac{2b}{3}=1$ $b=\frac32.$ Then $c=b=\frac32, \qquad a=\frac b2=\frac34.$

6. Compute

Now $\frac94+\frac94=\frac{18}{4}=\frac92=\frac{72}{16},$ so $a^2+b^2+c^2=\frac{9}{16}+\frac{72}{16}=\frac{81}{16}.$ Therefore, $16(a^2+b^2+c^2)=16\cdot \frac{81}{16}=81.$

7. Hence the required integer is $\boxed{81}.$

The derived answer matches the stored correct answer.