Indian Institute of Foreign Trade 2008 Solved Paper

It has been given that a, b, and c are in an arithmetic progression.
Let $\mathit{a}=\mathit{x}-\mathit{p},\mathit{b}=\mathit{x}$, and $\mathit{c}=\mathit{x}+\mathit{p}$
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.
$\frac{\mathit{a}+\mathit{b}+\mathit{c}}{3}\ge \sqrt[3]{\mathit{a}\mathit{b}\mathit{c}}$
$\frac{\mathit{a}+\mathit{b}+\mathit{c}}{3}\ge \sqrt[3]{4}$
$\frac{3\mathit{x}}{3}\ge \sqrt[3]{4}$
$\mathit{x}\ge \sqrt[3]{4}$
We know that $\mathit{x}=\mathit{b}$
Therefore,$\mathit{b}\ge \sqrt[3]{4}$ or $\mathit{b}\ge 2^{\frac{2}{3}}$
Therefore, option B is the right answer.