The random variable $$X$$ follows binomial distribution $$B(n, p)$$, for which the difference of the mean and the variance is 1.
If $$2P(X = 2) = 3P(X = 1)$$, then $$n^2P(X > 1)$$ is equal to

For binomial distribution B(n,p): Mean = np, Variance = npq where q = 1-p.

Given: np - npq = 1, so np(1-q) = 1, i.e., $$np^2 = 1$$ ... (1)

Given: $$2P(X=2) = 3P(X=1)$$

$$2\binom{n}{2}p^2q^{n-2} = 3\binom{n}{1}p q^{n-1}$$

$$2 \times \frac{n(n-1)}{2}p^2 q^{n-2} = 3np q^{n-1}$$

$$n(n-1)p^2 q^{n-2} = 3np q^{n-1}$$

$$(n-1)p = 3q = 3(1-p)$$

$$np - p = 3 - 3p$$

$$np = 3 - 2p$$ ... (2)

From (1): $$np^2 = 1$$, so $$p = \frac{1}{np}$$. From (2): $$np = 3-2p$$.

$$p = \frac{1}{3-2p}$$, so $$p(3-2p) = 1$$, $$3p - 2p^2 = 1$$, $$2p^2 - 3p + 1 = 0$$

$$(2p-1)(p-1) = 0$$, so $$p = 1/2$$ (since $$p \neq 1$$).

From (1): $$n \times (1/4) = 1$$, so $$n = 4$$. And $$q = 1/2$$.

$$P(X > 1) = 1 - P(X=0) - P(X=1)$$

$$P(X=0) = (1/2)^4 = 1/16$$

$$P(X=1) = 4(1/2)(1/2)^3 = 4/16 = 1/4$$

$$P(X>1) = 1 - 1/16 - 4/16 = 11/16$$

$$n^2 P(X>1) = 16 \times 11/16 = 11$$

This matches option 2: 11.