Answer
$0.88c$
Work Step by Step
We can find the speed of the rocket as follows:
$K.E=m_{\circ}c^2(\frac{1}{\sqrt{1-v^2/c^2}}-1)$
This can be rearranged as:
$v=c\sqrt{1-(\frac{1}{1+K.E/m_{c_{\circ}^2}})^2}$
We plug in the known values to obtain:
$v=c\sqrt{1-(\frac{1}{1+2.7\times 10^{23}/2.7\times 10^6(3.00\times 10^8)^2})^2}=0.88c$
