The random variable X follows binomial distribution B ( n, p ) ,...

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

$⇒ np - npq = 1$

$⇒ np ( 1 - q ) = 1$

$⇒ n{p}^{2} = 1$

Also given that,

$⇒ 2P ( X = 2 ) = 3P ( X = 1 )$

$⇒ 2 \cdot C2 n {p}^{2}{q}^{n - 2} = 3 \cdot C1 n p \cdot {q}^{n - 1}$

$⇒ 2 \cdot \dfrac{n \cdot ( n - 1 )}{2} \cdot p = 3 \cdot n \cdot q$

$⇒ ( n - 1 ) p = 3 ( 1 - p )$

$⇒ \left(\dfrac{1}{{p}^{2}} - 1\right)p = 3 ( 1 - p )$

$⇒ \dfrac{( 1 - p ) ( 1 + p )}{p} = 3 ( 1 - p )$

$⇒ 1 + p = 3p$

$⇒ p = \dfrac{1}{2}$

$∴ n = 4$.

Now,

${n}^{2}P ( x > 1 ) = {n}^{2} ( 1 - P ( x = 1 ) - P ( x = 0 ) )$$= 16\left(1 - C1 4 \cdot {\left(\dfrac{1}{2}\right)}^{4} - {\left(\dfrac{1}{2}\right)}^{4}\right) = 11$

Hence this is the correct option.