The random variable $X$ follows binomial distribution $B(n, p)$, for which the difference of the mean and the variance is $1$.

If $2 P(X=2)=3 P(X=1)$, then $n^{2}P(X>1)$ is equal to

Given that the difference between the mean and variance is $1$.

$\Rightarrow np-npq=1$

$\Rightarrow np(1-q)=1$

$\Rightarrow n{p}^2=1$

Also given that,

$\Rightarrow 2P(X=2)=3P(X=1)$

$\Rightarrow 2·{}^{n}C_2{p}^2{q}^{n-2}=3·{}^{n}C_{1}p·q^{n-1}$

$\Rightarrow 2·\frac{n·(n-1)}{2}·p=3·n·q$

$\Rightarrow (n-1)p=3(1-p)$

$\Rightarrow \left(\frac{1}{p^2}-1\right)p=3(1-p)$

$\Rightarrow \frac{(1-p)(1+p)}{p}=3(1-p)$

$\Rightarrow 1+p=3p$

$\Rightarrow p=\frac{1}{2}$

$\therefore n=4$.

Now,

$n^{2}P(x>1)=n^2(1-P(x=1)-P(x=0\left)\right)$$=16\left(1-{}^{4}C_1·{\left(\frac{1}{2}\right)}^4-{\left(\frac{1}{2}\right)}^4\right)=11$

Hence this is the correct option.