1. Given data

• Conductor lies along the $z$-axis from $z=-1.5$ to $z=1.5$ m, so its length is $L=3.0\,\text{m}$
• Current: $I=10.0\,\text{A}$ in direction $-\hat a_z$.
• Magnetic field: $\vec B=3.0\times 10^{-4} e^{-0.2x}\,\hat a_y\,\text{T}$
• Final position: $x=2.0$ m, $y=0$ m
• Time taken: $\Delta t=5\times 10^{-3}\,\text{s}$
• Motion is parallel to $x$-axis.

We assume the conductor starts at the $z$-axis, i.e. at $x=0$.

2. Magnetic force on the conductor

For a current-carrying conductor, $d\vec F= I\,d\vec l\times \vec B$

Here, $d\vec l = -\hat a_z\,dz$ because current flows in $-\hat a_z$ direction.

Also, $\vec B = 3.0\times10^{-4}e^{-0.2x}\hat a_y$

So, $d\vec F = I(-\hat a_z\,dz)\times \left(3.0\times10^{-4}e^{-0.2x}\hat a_y\right)$

Using $\hat a_z\times \hat a_y = -\hat a_x \quad \Rightarrow \quad (-\hat a_z)\times \hat a_y = +\hat a_x$

Therefore, $d\vec F = I\left(3.0\times10^{-4}e^{-0.2x}\right)dz\,\hat a_x$

Integrating over the full length $L=3$ m, $\vec F(x)= I L \left(3.0\times10^{-4}e^{-0.2x}\right)\hat a_x$

Substitute values: $F(x)=10\times 3\times 3.0\times10^{-4}e^{-0.2x}$ $F(x)=9.0\times10^{-3}e^{-0.2x}\,\text{N}$

This magnetic force is along $+x$.

3. Velocity of the conductor

It moves from $x=0$ to $x=2.0$ m in $5\times10^{-3}$ s at constant speed, so $v=\frac{2.0}{5\times10^{-3}}=400\,\text{m/s}$

4. Power required

Since magnetic force is in the same direction as motion, the external agent must apply an equal opposing force to maintain constant speed. The magnitude of mechanical power required at position $x$ is $P=Fv$

At the final position $x=2.0$ m, $F(2)=9.0\times10^{-3}e^{-0.4}$

Now, $e^{-0.4}\approx 0.6703$

Hence, $F(2)=9.0\times10^{-3}\times 0.6703 = 6.03\times10^{-3}\,\text{N}$

Therefore, $P=Fv=(6.03\times10^{-3})(400)$ $P\approx 2.41\,\text{W}$

This does not match any option.

5. Check if average power is intended

Because the magnetic field varies with $x$, sometimes such questions intend the average power over the motion.

Work done from $x=0$ to $x=2$: $W=\int_0^2 F(x)\,dx$ $W=\int_0^2 9.0\times10^{-3} e^{-0.2x}\,dx$ $W=9.0\times10^{-3}\left[\frac{-1}{0.2}e^{-0.2x}\right]_0^2$ $W=9.0\times10^{-3}\cdot 5\left(1-e^{-0.4}\right)$ $W=0.045(1-0.6703)$ $W\approx 0.01484\,\text{J}$

Average power: $P_{\text{avg}}=\frac{W}{\Delta t}=\frac{0.01484}{5\times10^{-3}}$ $P_{\text{avg}}\approx 2.97\,\text{W}$

This matches option B.

6. Conclusion

The option given in the answer key corresponds to the average power required during the motion, not the instantaneous power at $x=2$ m.

Thus the correct option is: $\boxed{\text{B }\; 2.97\,\text{W}}$