The correct answer is 3 m²s⁻¹.

We are given a particle of mass $m=1\text{kg}$ under the force:

$\overrightarrow{F}=-k(x\hat{i}+y\hat{j})$, with $k=1{\text{kg s}}^{-2}$

Since $m=1$, the equations of motion (by Newton's second law) are:

$\ddot{x}=-x$,    $\ddot{y}=-y$,    $\ddot{z}=0$

The key insight is to examine the quantity $L=x{v}_y-y{v}_x$, which is the z-component of the angular momentum per unit mass ($\overrightarrow{r}\times \overrightarrow{v}$).

Step 1: Show that L is a conserved quantity (constant in time).

Differentiate L with respect to time:

$\frac{\mathrm{dL}}{\mathrm{dt}}=x\frac{d{v}_y}{\mathrm{dt}}+v_x{v}_y-y\frac{d{v}_x}{\mathrm{dt}}-v_y{v}_x$

Substituting the accelerations $\frac{d{v}_x}{\mathrm{dt}}=-x$ and $\frac{d{v}_y}{\mathrm{dt}}=-y$:

$\frac{\mathrm{dL}}{\mathrm{dt}}=x(-y)+v_x{v}_y-y(-x)-v_y{v}_x=-xy+xy+v_x{v}_y-v_x{v}_y=0$

All terms cancel perfectly! This means $L=x{v}_y-y{v}_x$ is constant for all time. Its value does not change regardless of position or time, so we can evaluate it at $t=0$.

Step 2: Evaluate L at t = 0 using initial conditions.

At $t=0$, the given initial conditions are:

Position:   $x\left(0\right)=\frac{1}{\sqrt{2}}$ m,   $y\left(0\right)=\sqrt{2}$ m

Velocity:   $v_x\left(0\right)=-\sqrt{2}$ ms⁻¹,   $v_y\left(0\right)=\sqrt{2}$ ms⁻¹

Substituting into $L=x{v}_y-y{v}_x$:

$L=\frac{1}{\sqrt{2}}\times \sqrt{2}-\sqrt{2}\times (-\sqrt{2})$

$L=\frac{\sqrt{2}}{\sqrt{2}}+\sqrt{2}\times \sqrt{2}=1+2=3$

Step 3: Conclusion.

Since L is constant, this value of 3 holds for all future times and positions, including when $z=0.5\text{m}$. (The z-motion is entirely decoupled from x and y: with no z-force, $v_z=\sqrt{2}$ ms⁻¹ remains constant, and z increases uniformly to 0.5 m at $t=\frac{0.5}{\sqrt{2}}\text{s}$, but this has no effect on L.)

Therefore, when $z=0.5\text{m}$:

$(x{v}_y-y{v}_x)=3{\text{m}}^2{\text{s}}^{-1}$