Consider watching an alien space ship at Alpha Centauri (at 4.5 light years away) through a telescope from earth. This space ship turns towards us, and starts travelling toward us at what appears to be a velocity of exactly 2 million kilometers per second, when measuring the distance every hour.
The space ship will arrive here in about 8 months.
How fast is the space ship really travelling?
Suppose the spaceship is moving at speed $v$ and covers a distance $d$. The light emitted when the spaceship left reaches us at a time:
$$ t_0 = \frac{d}{c} $$
while the time the spaceship reaches us is:
$$ t_1 = d/v $$
The apparent velocity is distance divided by time so:
$$\begin{align} v_a &= \frac{d}{d/v - d/c} \\ &= \frac{vc}{c - v} \end{align}$$
Your question asks for the real velocity, so we rearrange to get:
$$ v = \frac{cv_a}{c + v_a} $$
If you substitute your value for $v_a$ you get $2.603×10^8\text{ m/s}$.
The answer to the apparent contradiction is that when you observe the spacecraft, it has long since left its starting point.
Suppose the spacecraft leaves at velocity $v=\beta c$ at time $t=0$ in the shared Earth / Alpha Centauri rest frame. Then the light emitted at that event will not reach earth until time $t=4\tfrac{6}{12}\:\mathrm{yr}$. If the spacecraft arrives on Earth eight months after that, at $t=5\tfrac{2}{12}\:\mathrm{yr}$, then its speed will be simply $$ v=\frac{\Delta x}{\Delta t}=\frac{4\tfrac{6}{12}}{5\tfrac{2}{12}}c=\tfrac{27}{31}c\approx0.87c=2.61\times10^8\:\mathrm{m}\:\mathrm{s}^{-1}, $$ which is of course slower than light, as the light got here first.
The spacecraft's apparent speed will, of course, be much faster, since we must observe it to transverse the $4.5 \:\mathrm{ly}$ in the eight months between our initial obervation and the spacecraft's arrival, which gives a speed of $2.61\times10^9\:\mathrm{m} \:\mathrm{s}^{-1} =6.75c$. John Rennie's answer provides a good explanation of how this number can be arrived at directly from the numbers above.
