y - x = z - y

x + z = 2y$\dots(1)$

xyz =4 $\dots(2)$

We know that, AM >= GM

Hence, $\frac{x+y+z}{3}$ >= $(xyz)^{\frac{1}{3}}$

From (1) and (2)

$\frac{3y}{3}$ >= $4^{\frac{1}{3}}$

y >= $2^{\frac{2}{3}}$

Hence,Option(B)$2^{\frac{2}{3}}$ is the correct choice.