Answer
$9.4\times 10^7Kg$
Work Step by Step
We know that
$\frac{m_1v_{1i}}{\sqrt{1-\frac{v_{1i}^2}{c^2}}}=\frac{(m_1+m_2)v}{\sqrt{1-\frac{v^2}{c^2}}}$
We plug in the known values to obtain:
$\frac{(8.2\times 10^6Kg)(0.5c)}{\sqrt{1-\frac{(0.50c)^2}{c^2}}}=\frac{(8.2\times 10^7Kkg)+m_2(0.26c)}{\sqrt{1-\frac{(0.260c)^2}{c^2}}}$
This simplifies to:
$m_2=9.4\times 10^7Kg$