1. We need to evaluate $p=\lim_{x\to 0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}}.$

2. This is of the form $1^{\infty}$, so take natural logarithm: $\ln p=\lim_{x\to 0}\frac{1}{x^2}\ln\left(\frac{\tan x}{x}\right).$

3. Use the standard expansion of $\tan x$ near $x=0$: $\tan x=x+\frac{x^3}{3}+O(x^5).$ Therefore, $\frac{\tan x}{x}=1+\frac{x^2}{3}+O(x^4).$

4. Now use $\ln(1+u)=u+O(u^2)\quad \text{for small }u.$ Here, $u=\frac{x^2}{3}+O(x^4).$ So, $\ln\left(\frac{\tan x}{x}\right)=\frac{x^2}{3}+O(x^4).$

5. Hence, $\ln p=\lim_{x\to 0}\frac{1}{x^2}\left(\frac{x^2}{3}+O(x^4)\right)=\frac{1}{3}.$ Thus, $p=e^{1/3}.$

6. Now compute: $96\ln p=96\cdot \frac{1}{3}=32.$

Therefore, the required integer is $\boxed{32}.$