Solution:

Given, standard deviation of first n natural numbers is 2.

Let us find the formula for the sum of first n natural numbers.

The sum of first n natural numbers is given by the formula:

S = n(n+1)/2

Let us find the formula for the variance of first n natural numbers.

Variance of first n natural numbers is given by the formula:

Variance = [∑(X - μ)²]/n

where X is the value of each observation, μ is the mean of the observations and n is the number of observations.

Let us find the mean of first n natural numbers.

Mean of first n natural numbers is given by the formula:

Mean = (sum of n natural numbers)/n

Mean = n(n+1)/2n

Mean = (n+1)/2

Let us substitute the values in the variance formula.

Variance = [∑(X - μ)²]/n

Variance = [∑(X - (n+1)/2)²]/n

Variance = [∑(X² - X(n+1)/2 + (n+1)²/4)]/n

Variance = (∑X² - (∑X(n+1)/2) + n(n+1)²/4)/n

Variance = (∑X² - n(n+1)²/4)/n

Let us find the formula for the sum of squares of first n natural numbers.

The sum of squares of first n natural numbers is given by the formula:

∑X² = n(n+1)(2n+1)/6

Let us substitute the value of ∑X² in the variance formula.

Variance = (∑X² - n(n+1)²/4)/n

Variance = [(n(n+1)(2n+1)/6) - n(n+1)²/4]/n

Variance = (n+1)(n-1)/12

Let us substitute the given value of standard deviation in terms of variance.

Standard deviation = √Variance

2 = √[(n+1)(n-1)/12]

4 = (n+1)(n-1)/12

48 = (n+1)(n-1)

48 = n² - 1

n² = 49

n = 7

Therefore, the value of n is 7.

Hence, option 'B' is the correct answer.