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.