In the fall of 2002, a group of scientists at Los Alamos National Laboratory determined that the critical mass of neptunium-237 is about 60

I just want to think this through more than Alan needed to.     

$$Volume\;of\:sphere=\frac{4}{3}\pi r^3$$         $$Need\; to \;find\; r\;\;cm$$ 

$$Density=\dfrac{19.5g}{cm^3}$$     

  $$If\; there \;is \;\dfrac{19.5g}{cm^3} \qquad Then\; it \;would\; also\; be\; true \;to\; say \;that\; there\; are \; \dfrac{cm^3}{19.5g}$$

$$mass=60kg=60000g$$

so

$$\dfrac{cm^3}{19.5g}\times 60000g=3076.923cm^3 \qquad \mbox{The grams cancel out}$$

 

$$\begin{array}{rll}
3076.923&=&\frac{4}{3}\pi r^3\\\\
3076.923*3/(4\pi) &=& r^3\\\\
r&=&\sqrt[3]{3076.923*3/(4\pi) }\\\\
r&=&9.023cm
\end{array}$$

 

$${\left({\frac{{\mathtt{3\,076.923}}{\mathtt{\,\times\,}}{\mathtt{3}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = {\mathtt{9.022\: \!827\: \!882\: \!792\: \!352}}$$

Density is mass per unit volume, so volume = mass/density.  Volume of a sphere is (4/3)pi*r³, where r is the radius.  Put these together to get 

r = (3*mass/[4pi*density])^1/3.

$${\mathtt{r}} = {\left({\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{60\,000}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{19.5}}\right)}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} \Rightarrow {\mathtt{r}} = {\mathtt{9.022\: \!827\: \!957\: \!982\: \!585}}$$

r ≈ 9 cm

(Note that I've expressed the mass as 60000 grams because the density is given as grams/cm³)