$\large\frac{g_2}{g_1}=\frac{{R_1}^2}{{R_2}^2}$

$R_1=R$

$R_2=2R$

$\large\frac{g_2}{g_1}=\frac{R^2}{4R^2}$

$\qquad=\large\frac{1}{4}$

$T= 2 \pi \sqrt {\large\frac{l}{g}}$

$\large\frac{T_2}{T_1}=\sqrt {\large\frac{g_1}{g_2}}$

$\qquad=\sqrt 4=2$

Hence d is the correct answer.