Ionic Radius of Iodine in High Temperature Form
Considering by cubic lattice you mean a Simple Cubic structure, you could do the following:


Reference image1 
Reference image2


(could not put it here because of copyright)

In a cubic lattice the atoms touch each other along the edges(see image1 and image2), hence the edge length/parameter will be equal to twice the radius (see image2). 
$$\therefore\ \ce{Edge\ Length(a)=2\times radius(r)}$$
$$\therefore\ a \ = \ 2\times r$$

This method can be adopted for any lattice, you only need to find the axis along which the atoms touch each other(image1 bottom row).  



If they exhibited an fcc close packing structure, you would notice that the atoms touch each other along a face-diagonal. 

$$
\begin{align}
\therefore\ce{Length\ of\ face\ diagonal&amp;\ =\ radius\ of\ top\ left\ atom +\\
                                        &amp;\quad\ diameter\ of\ atom\ at\ face-center + \\
                                        &amp;\quad\ radius\ of\ bottom\ left\ atom}\\
\therefore\ce{Length\ of\ face\ diagonal&amp;\ =\ r\ + ( 2\times r ) + r}\\
\therefore\ce{\sqrt{2}\times a&amp;\ =\ 4\times r}\\
\therefore\ce{a&amp;\ = 2\sqrt{2}\times r}
\end{align}
$$



Similarly for bcc structure the atoms touch each other along the body-diagonal 
$$\therefore\ce{\sqrt{3}\times a\ =\ 4\times r}$$