1. Given data

• Mass of body: $m = 2\,\text{kg}$
• Initial velocity: $u = 0$
• Power is constant: $P$
• Time: $t = 4\,\text{s}$
• Displacement in $4\,\text{s}$ is given as $s = \frac{1}{3}\alpha^2\sqrt{P}\,\text{m}$

We need to find $\alpha$.

2. Use relation between power and velocity

For a body moving under constant power, $P = Fv$

Also, $F = m\frac{dv}{dt}$

So, $P = mv\frac{dv}{dt}$

Rearranging, $v\,dv = \frac{P}{m}\,dt$

Integrating from $v=0$ at $t=0$ to velocity $v$ at time $t$: $\int_0^v v\,dv = \frac{P}{m}\int_0^t dt$

$\frac{v^2}{2} = \frac{P}{m}t$

Hence, $v = \sqrt{\frac{2Pt}{m}}$

3. Find displacement

Since $v = \frac{ds}{dt} = \sqrt{\frac{2Pt}{m}}$

So, $ds = \sqrt{\frac{2P}{m}}\, t^{1/2} dt$

Integrating from $0$ to $t$: $s = \sqrt{\frac{2P}{m}}\int_0^t t^{1/2}dt$

$s = \sqrt{\frac{2P}{m}}\left[\frac{2}{3}t^{3/2}\right]_0^t$

Thus, $s = \frac{2}{3}\sqrt{\frac{2P}{m}}\,t^{3/2}$

4. Substitute $m=2\,\text{kg}$ and $t=4\,\text{s}$

Since $m=2$, $\sqrt{\frac{2P}{m}} = \sqrt{\frac{2P}{2}} = \sqrt{P}$

Also, $t^{3/2} = 4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$

Therefore, $s = \frac{2}{3}\sqrt{P}\cdot 8 = \frac{16}{3}\sqrt{P}$

This is given as $s = \frac{1}{3}\alpha^2\sqrt{P}$

Equating, $\frac{1}{3}\alpha^2\sqrt{P} = \frac{16}{3}\sqrt{P}$

Cancel $\frac{1}{3}\sqrt{P}$ from both sides: $\alpha^2 = 16$

Hence, $\alpha = 4$

5. Final answer

$\boxed{4}$