The relative basics strenghts of $N H_{3}, C H_{3} N H_{2}$ and $N F_{3}$ are in the order:

C H_{3} N H_{2} > N H_{3} > N F_{3}

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N F_{3} > C H_{3} N H_{2} > N H_{3}

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N H_{3} > C H_{3} N H_{2} > N F_{3}

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N F_{3} > C H_{3} N H_{2} = N H_{3}

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Solution

The correct option is A $C H_{3} N H_{2} > N H_{3} > N F_{3}$
$C H_{3} -$ group is electron donating group.
$F$ is electron withdrawing group.
So $C H_{3} N H_{2}$ is more basic. $N H_{3}$ is more basic than $N F_{3}$ because $H$ is not a withdrawing group. $N F_{3}$ is less basic.
$∴$ Basic nature order= $C H_{3} N H_{2} > N H_{3} > N F_{3}$

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