From stationary, we can say that

$$Pt = \frac{1}{2}mv^2 \quad \therefore \quad v = \sqrt{\frac{2Pt}{m}} \quad \therefore \quad \frac{\mathrm{d}v}{\mathrm{d}t} = \sqrt{\frac{P}{2mt}}$$

In general, $Pt = \frac{1}{2}m\left( v_2^2-v_1^2 \right)$.

If $v_2 = v_1 + 1$, then

$$Pt = \frac{1}{2}m\left( (v_1+1)^2-v_1^2 \right) = \frac{1}{2}m(2v_1 + 1)$$

If $t = 1 \textrm{ day } = 86400 \mathrm{\ s}$

$$v_1 = \frac{Pt}{m} - \frac{1}{2} = \frac{86400P}{m} - \frac{1}{2}$$

$$\frac{86400P}{m} - \frac{1}{2} = \sqrt{\frac{2Pt}{m}} \Rightarrow \frac{2Pt}{m} = \left( \frac{86400P}{m} - \frac{1}{2} \right) ^2$$

$$\therefore t = \frac{m}{2P} \left( \frac{86400P}{m} - \frac{1}{2} \right) ^2$$

after which it will take one day to accelerate by $1 \mathrm{\ m\ s^{-1}}$.

We are told that $\frac{P}{m} = 500 \mathrm{\ W\ kg^{-1}}$

$$\therefore t = \frac{1}{1000} \left( 43200000 - \frac{1}{2} \right) ^2 = 1.86624 \times 10^{12} \mathrm{\ s} = 21.6 \textrm{ million days}$$

$v = 4.32 \times 10^7 \mathrm{\ m\ s^{-1}}$, which is around $15\%$ of the speed of light.