#### Answer

The third piece moves forward with a speed of $1.5\times 10^7~m/s$.

#### Work Step by Step

We can find the mass $m_3$ of the third piece as;
$m_1+m_2+m_3 = m_0$
$m_3 = m_0-m_1-m_2$
$m_3 = (2.0\times 10^6~kg)-(5.0\times 10^5~kg)-(8.0\times 10^5~kg)$
$m_3 = 7.0\times 10^5~kg$
We can use conservation of momentum to find the velocity $v_3$ of the third piece after the explosion.
$p_f=p_0$
$m_1~v_1+m_2~v_2+m_3~v_3 = m_0~v_0$
$v_3 = \frac{m_0~v_0-m_1~v_1-m_2~v_2}{m_3}$
$v_3 = \frac{(2.0\times 10^6~kg)(5.0\times 10^6~m/s)-(5.0\times 10^5~kg)(-2.0\times 10^6~m/s)-(8.0\times 10^5~kg)(1.0\times 10^6~m/s)}{7.0\times 10^5~kg}$
$v_3 = 1.5\times 10^7~m/s$
The third piece moves forward with a speed of $1.5\times 10^7~m/s$.