How fast?

First we need to re-arrange the equation to make $v$ the subject since that is what we're calculating. It's a bit long winded so you can skip ahead if you want.

$E^2 = \left(\frac{m_0c^2}{\sqrt{1 - \frac{v^2}{c^2}}}\right)^2 + \left(\frac{m_0vc}{\sqrt{1 - \frac{v^2}{c^2}}}\right)^2$

$E^2 = \frac{m_0^2c^4}{1 - \frac{v^2}{c^2}} + \frac{m_0^2v^2c^2}{1 - \frac{v^2}{c^2}}$

$E^2 = \frac{m_0^2c^4 + m_0^2v^2c^2}{1 - \frac{v^2}{c^2}}$

$E^2\left(1 - \frac{v^2}{c^2}\right) = m_0^2c^4 + m_0^2v^2c^2$

$E^2 - \frac{E^2v^2}{c^2} = m_0^2c^4 + m_0^2v^2c^2$

$E^2 = m_0^2c^4 + m_0^2v^2c^2 + \frac{E^2v^2}{c^2}$

$E^2 - m_0^2c^4 = v^2\left(m_0^2c^2 + \frac{E^2}{c^2}\right)$

$v^2 = \frac{E^2 - m_0^2c^4}{m_0^2c^2 + \frac{E^2}{c^2}}$

$v = \sqrt{\frac{E^2 - m_0^2c^4}{m_0^2c^2 + \frac{E^2}{c^2}}}$

This is much easier to work with now, however we only know $m_0$ so we can't work out $v$ just yet. We're missing a value for $E$ , luckily we've been given a second mass to work with. Why does this help? Because we now have the required components to substitute in $E = mc^2$ .

$v = \sqrt{\frac{(mc^2)^2 - m_0^2c^4}{m_0^2c^2 + \frac{(mc^2)^2}{c^2}}}$

Simplifying this down gives us:

$v = \sqrt{\frac{m^2c^4 - m_0^2c^4}{m_0^2c^2 + \frac{m^2c^4}{c^2}}}$

$v = \sqrt{\frac{c^4(m^2 - m_0^2)}{m_0^2c^2 + m^2c^2}}$

$v = \sqrt{\frac{c^4(m^2 - m_0^2)}{c^2(m^2+ m_0^2)}}$

$v = \sqrt{\frac{c^2(m^2 - m_0^2)}{m^2+ m_0^2}}$

$v = c\sqrt{\frac{m^2 - m_0^2}{m^2+ m_0^2}}$

Now all we have to do is substitute in values for $m_0$ and $m$ , $10^{12}$ and $5.5 \cdot 10^{12}$ respectively.

$v = c\sqrt{\frac{(5.5\cdot10^{12})^2 - (10^{12})^2}{(5.5\cdot10^{12})^2 + (10^{12})^2}}$

$v = c\sqrt{\frac{30.25\cdot10^{24} - 10^{24}}{30.25\cdot10^{24} + 10^{24}}}$

$v = c\sqrt{\frac{29.25 \cdot 10^{24}}{31.25 \cdot 10^{24}}}$

$v = c\sqrt{\frac{29.25}{31.25}}$

$v = c\sqrt{0.936}$

$v = 0.96747092...c$

Now that we have a value we just need to round it to $6\text{.d.p}$ .

$v = 0.967471c$

The question requires that we give just the numerical part of the answer so the answer is $\boxed{0.967471}$