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(B) In the $C_3$ cycle (Calvin cycle),the fixation of $6$ molecules of $CO_2$ is required to produce one molecule of glucose $(C_6H_{12}O_6)$.For the

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$18$ and $12$

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(B) In the $C_3$ cycle (Calvin cycle),the fixation of $6$ molecules of $CO_2$ is required to produce one molecule of glucose $(C_6H_{12}O_6)$.For the fixation of each molecule of $CO_2$,$3$ $ATP$ and $2$ $NADPH_2$ are consumed.Therefore,for $6$ molecules of $CO_2$:$ATP$ required = $6 	imes 3 = 18$ $ATP$.$NADPH_2$ required = $6 	imes 2 = 12$ $NADPH_2$.Thus,the correct answer is $18$ $ATP$ and $12$ $NADPH_2$.

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