There is a number x such that

(a) Let $x=\sqrt{5}$.

$x^{2}=(\sqrt{5})^{2}=5$, which is a rational number.

(b) Let $x=\sqrt{2}$.

$x^{2}=(\sqrt{2})^{2}=2$, which is a rational number.

(c) Let $x=\sqrt[3]{2}$

$x^{2}=(\sqrt[3]{2})^{2}=(2)^{\frac{2}{3}}$, which is an irrational number.

$x^{4}=(\sqrt[3]{2})^{4}=(2)^{\frac{4}{3}}$, which is also an irrational number.

(d) Let $x=\sqrt[4]{2}$

$x^{2}=(\sqrt[4]{2})^{2}=(2)^{\frac{2}{4}}=(2)^{\frac{1}{2}}$, which is an irrational number.

$x^{4}=(\sqrt[4]{2})^{4}=(2)^{\frac{4}{4}}=2$, which is a rational number.

$\therefore x$ can be $\sqrt[4]{2}$

Hence, the correct option is (d).