For an approximate Potential due to the magnetic dipole, we have:
$\vec{A}=\frac{\mu_{0}}{4 \pi}\frac{\vec{m} \times \hat{r}}{r^{2}}$
Now, for $\vec{m}=Iw^{2}\hat{z}$,
we get$ \vec{A}=\frac{\mu_{0}}{4 \pi r^{2}}\left [ Iw^{2} sin \theta \right ]$
However, my solution indicates $\vec{A}=\frac{\mu_{0}}{4 \pi r^{2}}\left [ Iw^{2} sin \theta \right ]\hat{\phi}$
I'm frustrated at being stuck with these simple directions for hours.
Would someone kindly help me?
The use of both $\hat z$ and $\hat r$ in the question tends to give the impression that you are working in cylindrical coordinates, so that $\hat r$ is perpendicular to $\hat z$ at all points in space. But that interpretation of $\hat r$ contradicts the formula $$\vec A = \frac{\mu_0}{4 \pi}\frac{\vec m \times \hat r}{r^2} \tag1$$ for magnetic potential; the formula is correct only when $\vec r$ is the complete position vector in three-dimensional space and $\hat r = \frac{\vec r}{r}$. In general, in order to use Equation $(1)$ for the magnetic dipole potential, $\hat r$ is never perpendicular to $\hat z$ except at points in the $x,y$ plane.
Rather than cylindrical basis vectors, what you have is a mix of spherical and Cartesian basic vectors. The spherical basis vectors $\hat r$ and $\hat \phi$ are illustrated in the following figure from the document about spherical coordinates linked here:
Your $\hat z$ unit vector is not shown in the figure but it points in the direction of the positive $z$ axis, just as it would if we were drawing the basis vectors of the Cartesian coordinates. So the angle between $\hat z$ and $\hat r$ is $\theta$ as shown in the figure, and $$ \hat z \times \hat r = (\sin \theta) \,\hat \phi. $$
The reason the cross product is in the direction $\hat \phi$ is because $\hat\phi$ is perpendicular to the plane that contains $\hat z$ and $\hat r$. The sign (that is, why it is $(\sin \theta) \,\hat \phi$ instead of $-(\sin \theta) \,\hat \phi$) is due to the right-hand rule for the cross-product, which works the same in every coordinate system because this right-hand rule is defined on an arbitrary pair of vectors, not on the Cartesian axes. Take your right hand and make the hitchhiker's "thumbing a ride" sign with it:
If you stick your thumb in the direction of the cross-product then your fingers will be curled around that line in the direction that takes you from the first input vector of the cross product to the second input vector without going more than $\pi$ radians ($180$ degrees) around. If you find your fingers are curling from the second vector toward the first vector instead, you stuck your thumb in the wrong direction and must reverse it.