Question 1187361
Strange problem.
{{{1trillion=10^12}}} in the USA
(Between different countries, we do not agree on the definitions of billion, trillion, trapezoid, pavement, pants, hood, trunk, engine, and other terms).
{{{1yard=3feet=36inches}}}
A cube with {{{5in}}} edges has an outside surface area of
{{{6(5in)(5in)=highlight(150in^2)}}}
 
If that outside surface area was divided into
{{{30trillion=30*10^12}}} pieces of equal area,
each piece would have an area of
{{{(150in^2/(30*10^12))(1yard/36in)(1yard/36in)}}}={{{150*10^(-12)/(30*36*36)}}}{{{yard^2}}}={{{cross(30)*5*10^3*10^(-3)*10^(-12)/(cross(30)*1296)}}}{{{yard^2}}}
={{{5000*10^(-15)/1296}}}{{{yard^2}}}={{{highlight(3.86*10^(-15))}}}{{{yard^2}}}

Hopefully, that is an acceptable answer, and there is no objection to having 3 significant figures in the result.
Otherwise, maybe {{{highlight(3.9*10^(-15))}}}{{{yard^2}}} , with 2 significant figures, may be the expected answer, since 30 trillion appears to have 2 significant figures.