So what you do is to realize that you can replace $(k_x,k_y,k_z)\cdot(x,y,z)$ by $\kappa r \cos(\theta)$, where $\kappa = \sqrt{k_x^2 + k_y^2 + k_z^2}$ and $r = \sqrt{x^2+y^2+z^2}$ and $\theta$ is the angle between $(k_x,k_y,k_z)$ and $(x,y,z)$. Also, you have that the Fourier transform of the constant function is the Dirac delta function, so we can consider $1 - H(\sqrt{k_x^2 + k_y^2 + k_z^2} - k_0) = H(k_0 - \sqrt{k_x^2 + k_y^2 + k_z^2})$ instead. Then using spherical coordinates as shown in the first picture on this website https://en.wikipedia.org/wiki/Spherical_coordinate_system, making $(k_x,k_y,k_z)$ parallel to the $z$-axis in this same picture, and replace $r$ with $\kappa$:
$$ \int \frac{d k_x d k_y d k_z}{(2 \pi)^3} H(k_0 - \sqrt{k_x^2 + k_y^2 + k_z^2}) \exp{(- \text{i} (k_x, k_y, k_z) \cdot (x , y ,z))} $$
is equal to
$$ \int_{\theta=0}^{\pi}\int_{\kappa=0}^\infty \int_{\varphi=0}^{2\pi} \frac{\kappa^2 \sin\theta \, d\kappa \, d\theta \, d\varphi}{(2\pi)^3} H(k_0-\kappa) e^{-i\kappa r \cos\theta} $$
which equals
$$ \int_{\kappa=0}^{k_0} \int_{\theta=0}^{\pi} \frac{2 \pi \kappa^2 \sin\theta \, d\kappa \, d\theta}{(2\pi)^3} e^{-i\kappa r \cos\theta} $$
Being lazy, we plug this into Wolfram Alpha: http://www.wolframalpha.com/input/?i=Integrate%5B2+Pi+k%5E2+Sin%5Bt%5D+Exp%5B-I+k+r+Cos%5Bt%5D%5D%2F%282+Pi%29%5E3%2C+%7Bk%2C+0%2C+k0%7D%2C+%7Bt%2C+0%2C+Pi%7D%5D
to get
$$ \frac{ \sin (k_0 r
) - k_0 r \cos (k_0 r )}{2 \pi ^2 r ^3} .$$
