Q.Biot-Savart law indicates that the moving electrons (velocity $\vec{v}$) produce a magnetic field $\vec{B}$ such that

This is a Biot-Savart question about the FIELD a moving charge creates, not the force on it.

A charge $q$ moving with velocity $\vec{v}$ produces, at a point whose position vector from the charge is $\vec{r}$ (unit vector $\hat{r}$), a magnetic field

Because $\vec{B}$ comes from the cross product $\vec{v}\times\hat{r}$, it is perpendicular to both $\vec{v}$ and $\vec{r}$. In particular $\vec{B}\perp\vec{v}$.

What the question asks

It asks about the magnetic field a moving electron creates (the Biot-Savart law), not the force a field exerts on a current.

The Biot-Savart field of a moving charge

For a charge $q$ moving with velocity $\vec{v}$, the field at a point with position vector $\vec{r}$ (unit vector $\hat{r}$) from the charge is

1. Direction. A cross product $\vec{v}\times\hat{r}$ is perpendicular to both vectors that build it. Hence $\vec{B}$ is perpendicular to $\vec{v}$ and to $\vec{r}$. So $\vec{B}\perp\vec{v}$.

2. Magnitude. $B = \dfrac{\mu_0}{4\pi}\dfrac{q\,v\sin\theta}{r^2}$, where $\theta$ is the angle between $\vec{v}$ and $\vec{r}$. This is an inverse-square law in $r$, not inverse-cube.

3. Rejecting the other choices. $\vec{B}$ is not parallel to $\vec{v}$ (a cross product is never parallel to its factors), and it does not lie along the line joining the charge to the point (that line is $\hat{r}$, to which $\vec{B}$ is also perpendicular).

Method: Reading Off Field Direction/Magnitude Directly from the Biot-Savart Cross Product

For any "what does the Biot-Savart law say about the field of a moving/current-carrying source" question, the entire answer comes from the geometric properties of a cross product -- you rarely need to compute an actual number.

Steps

Step 1: Write the Biot-Savart field in cross-product form

For a moving point charge $q$ (or a current element $I\,d\vec l$), the field at a point displaced by $\hat r$ is:

Step 2: Use the cross-product rule to fix the direction

A cross product $\vec a\times\vec b$ is always perpendicular to both $\vec a$ and $\vec b$ -- never parallel to either, and never lying along a third, unrelated direction. So here $\vec B$ must be perpendicular to $\vec v$ (the source's velocity/current direction) and perpendicular to $\hat r$ (the line to the field point) simultaneously. This single fact eliminates "parallel to $\vec v$" and "along the line joining source and point" options immediately.

Step 3: Use the $1/r^2$ prefactor to fix the distance dependence

The explicit $r^2$ in the denominator makes this an inverse-square law, exactly like Coulomb's law for $\vec E$ -- never inverse-cube, inverse-linear, etc. Any option proposing a different power of $r$ can be rejected on sight from the formula's structure.

Applying to this problem: with $\vec v$ as the electron's velocity, $\vec B\perp\vec v$ follows immediately from Step 2 -- no vector algebra with actual numbers is needed to answer a qualitative direction/magnitude-dependence question like this one.