The correct answer is $\boxed{\frac{{\sqrt 3 }}{2} + {\text{j}}\frac{1}{2}}$.
To solve for the roots of the equation $x^3 = j$, we can use the cubic formula:
$$x = \dfrac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$
where $a = 1$, $b = 0$, and $c = -j$.
Substituting these values into the formula, we get:
$$x = \dfrac{0 \pm \sqrt{0^2 – 4 \cdot 1 \cdot -j}}{2 \cdot 1}$$
$$x = \dfrac{0 \pm \sqrt{4j}}{2}$$
$$x = \dfrac{0 \pm 2\sqrt{j}}{2}$$
$$x = \pm \sqrt{j}$$
Since $j$ is the positive square root of $-1$, then $\sqrt{j} = j$. Therefore, the roots of the equation $x^3 = j$ are $x = \pm j$.
We can also solve for the roots by inspection. Note that $j$ is a complex number with magnitude $1$ and angle $\frac{\pi}{3}$. Therefore, the roots of the equation $x^3 = j$ must be complex numbers with magnitude $1$ and angles $\frac{\pi}{3}$, $\frac{2\pi}{3}$, and $\frac{4\pi}{3}$. The only complex number with magnitude $1$ and angle $\frac{\pi}{3}$ is $\frac{{\sqrt 3 }}{2} + {\text{j}}\frac{1}{2}$. Therefore, $\boxed{\frac{{\sqrt 3 }}{2} + {\text{j}}\frac{1}{2}}$ is one of the roots of the equation $x^3 = j$.
The other two roots of the equation $x^3 = j$ are $-\frac{{\sqrt 3 }}{2} – {\text{j}}\frac{1}{2}$ and $\frac{{\sqrt 3 }}{2} – {\text{j}}\frac{1}{2}$.