Question 1027858
So assume that the side was originally, {{{S}}}.
Now it would be {{{S+r/100}}}
The volume originally was 
{{{V[1]=S^3}}}
and the new volume,
{{{V[2]=(S+r/100)^3}}}}
{{{V[2]=S^3+(3/100)S^2r+(3/10000)Sr^2+(1/1000000)r^3}}}
We can approximate by chopping off the last two terms,
{{{V[2a]=S^3+(3/100)S^2r}}}
So then comparing,
{{{V[2a]/V[1]=(S^3+(3/100)S^2r)/S^3}}}
{{{V[2a]/V[1]=1+(3/100)(r/S)}}}
Multiply by 100 to make it a percentage,
{{{V[2a]/V[1]=100+3(r/S)}}}
Subtract 100 to find the percentage change,
{{{PC=3(r/S)}}}