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

$$V = \frac{4}{3}\pi \left(\frac{5000}{2}\right)^3= 6.54\times 10^{10} \mathrm{m}^3$$

The density $\rho$ of the comet in $\textrm{kg m}^{-3}$ is

$$300 < \rho < 700\;.$$

Using $m = \rho V$ we can therefore estimate the mass $m$ of the comet, in $\mathrm{kg}$, to be bounded as

$$1.96\times 10^{13}< m< 4.58\times 10^{13}\;.$$

The momentum of the comet, measured in $\textrm{kg m s}^{-1}$, 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 $\textrm{m s}^{-1}$, 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 $2.3\times 10^{-7}\textrm{m s}^{-1}$.

In every-day units this is about $8.4\times 10^{-7}\textrm{km h}^{-1}$, which is tiny. (Note: the earth's velocity around the sun is around $30\textrm{ km s}^{-1}$.)