Answer
a. $1\%$
b. $3\%$
Work Step by Step
a. For a surface area $\frac{dS}{S}\leq2\%$, we have $ds\leq0.02S$. With $S=6a^2$, we have $dS=12a(da)$ and $da=\frac{dS}{12a}$ where $a$ is the edge of the cube. We have $\frac{da}{a}=\frac{ds}{12a^2}=\frac{ds}{2S}\leq1\%$
b. Given $\frac{da}{a}\leq1\%$ from above, with $V=a^3$, we have $dV=3a^2da$ and $\frac{dV}{V}=\frac{3a^2da}{a^3}=3\frac{da}{a}\leq3\%$
