The number of ways in which we can choose a committee from four men and six women so that

Number of men = 4 

Number of women = 6    (given) 

Given condition says, committee include at-least two man and exactly twice as many women as men . 

So, we can select either 2 men and 4 women or 3 men and 6 women. 

$\therefore$ Required number of ways of forming committee is ${ }^{4} C_{2} \times{ }^{6} C_{4}+{ }^{4} C_{3} \times{ }^{6} C_{6}$

i. e $\frac{4 !}{2 ! 2 !} \times \frac{6 !}{4 ! 2 !}+\frac{4 !}{3 !} \times 1$

$=\frac{4 \times 3}{2} \times \frac{6 \times 5}{2}+4$

$=6 \times 15+4$

$=94$

Hence, the correct answer is option A.