If the positive real numbers a, b and c are in Arithmetic Progression, such that abc = 4, then minimum possible value of b is:

It has been given that a, b, and c are in an arithmetic progression.
Let a = x-p, b = x, and c = x+p
We know that a, b, and c are real numbers.
Therefore, the arithmetic mean of a,b,c should be greater than or equal to the geometric mean.
$$\frac{a+b+c}{3} \geq \sqrt[3]{abc}$$
$$\frac{a+b+c}{3} \geq \sqrt[3]{4}$$
$$\frac{3x}{3}\geq\sqrt[3]{4}$$

$$x\geq\sqrt[3]{4}$$
We know that $$x$$ = $$b$$.
Therefore,$$b\geq\sqrt[3]{4}$$or $$b\geq 2^{\frac{2}{3}}$$
Therefore, option B is the right answer.