We know that

$T^2 \;\alpha \; r^3$

$=> T \; \alpha \; r^{3/2}$

$ \large\frac{T_{\Large mars}}{T_{\Large earth}}=\bigg(\Large\frac{r _{mars}}{r_{earth}}\bigg)^{3/2}$

$\qquad= (1.524)^{3/2}$

$\qquad= 1.88 $

$T_{\large earth}=1 year$

$\therefore T_{\large mars}= 1.88 \;earth \;years$

Hence b is the correct answer.