Total jurors = $15$
Men = $\frac{2}{3}(15) = 10$
Women = $15 - 10 = 5$

Jury = $12$
Men (at-least) = $\frac{2}{3}(12) = 8$

Case I: 8M4W
possible selections = $(^{10}C_8)(^5C_4) = 225$

Case II: 9M3W
possible selections = $(^{10}C_9)(^5C_3) = 100$

Case III: 10M2W
possible selections = $(^{10}C_{10})(^5C_2) = 10$

Total possibilities of selecting $12$ members from $15$ = $^{15}C_{12} = 455$

Required probability = $\frac{(225 + 100 + 10)}{455} = \frac{335}{455} = \frac{67}{91}$

Hence, option D