$T=2 \pi \sqrt {\large\frac{l}{g}}$
$g=\large\frac{4 \pi^2l}{T^2}$
$\large \frac {\Delta g}{g}=\frac {\Delta l}{l}+\frac{ 2 \Delta T}{T}$
$\qquad=\bigg(\large\frac{0.1}{50}+\frac{2 \times 0.01}{2.00}\bigg)$$ \times 100$
$\qquad=1.2 \%$
Hence d is the correct answer.