This problem involves momentum which we calculate through the formula $p=mv$. We need to use standard units (or, at least, the same units) throughout. We will keep a sensible number of places for accurary.

Volume of comet

$V = \frac{4}{3}\pi \left(\frac{5000}{2}\right)^3=   6.54\times 10^{10} \mbox{m}^3$
 
The density $\rho$ of the comet in kg m$^{-3}$ is

$300 < \rho < 700$ 
 
Using $m = \rho V$ we can therefore estimate the mass $m$ of the comet, in kg, to be bounded as
 
$1.96\times 10^{13}< m< 4.58\times 10^{13}$

The momentum of the comet, measured in kg m s$^{-1}$, can now be estimated as
 
$1.18 \times 10^{18}<   p < 2.75 \times 10^{18}$
 
If we assume that all of this momentum is transferred to the earth on impact then the velocity change $\Delta V$, measured in m s$^{-1}$, that this would impart to the earth is found by $\Delta v = p/M$ where $M$ is the mass of the earth.
 
$1.97 \times 10^{-7} <  \Delta v < 4.60 \times 10^{-7}$
 
Taking the centre of the range we can estimate a velocity change for the earth of about $2.3\times 10^{-7}$ ms$^{-1}$. 
 
In every-day units this is about $8.4\times 10^{-7}$ km h$^{-1}$, which is tiny. (Note: the earth's velocity around the sun is around $30$ km s$^{-1}$)