Let S denote area,
S=$\frac{1}{2}bcsinA$
Given $\frac{\Delta A}{A}=0.01$
A=$\frac{\pi}{6} \implies \Delta A=0.01 \times \frac{\pi}{6} $
$dS=\frac{1}{2}d(bcsinA)=\frac{1}{2}(csinAdb+bsinAdc+bccosAdA)$
$dS=\frac{1}{2}d(bcsinA)=\frac{1}{2}(cdb+bdc+bccotAdA)$
$dS=\frac{1}{2}d(bcsinA)=\frac{1}{2}(\frac{db}{b}+\frac{dc}{c}+cotAdA)$
Now we have :
$\frac{dS}{S}=\frac{db}{b}+\frac{dc}{c}+\frac{dA}{tanA}$
$\implies \frac{ds}{S}=\frac{0.01 \times \frac{\pi}{6}}{tan30}=0.01 \times \sqrt 3 \times \frac{\pi}{6}$. To match an option maybe there is some approximation.