$(m)_{10}$=$(p)_{2}$= $(q)_{3}$

$\because$ p is in base 2 $\Rightarrow$ its digit can be only combination of 1 and 0.

$\because$ q is in base 3 $\Rightarrow$ its digit can be only combination of 0,1 and 2.

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$(p)_{10}$- $(q)_{10}$= $(990)_{10}$

In this the last digit of p and q should be same in order to get 0.

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**Option B**

9 $\leq$ m $\leq$ 13

$(13)_{10}$=$(1101)_{2}$= $(111)_{3}$

$\because$ last digit of 1101 and 111 is same and 1101-111 = 990

$\therefore$This could be the answer.

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**Option A**

m$\geq$ 14

$\because$ m= 13 is satisfying

$\therefore$This could not be the answer.

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**Option C**

6 $\leq$ m $\leq$ 8

$(8)_{10}$=$(1000)_{2}$= $(22)_{3}$

$\because$ last digit of 1000 and 22 is not same this could not be the answer.

$\because$ the remaining numbers(m=6,7) when converted to p will be less than 1000 and would be a 3 digit number which would be combination of 1 and 0 so the difference will never be $\geq$ 990 in this case.

$\therefore$ This will never be the answer.

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**Option D**

m$<$6

$\because$ the remaining numbers(m<6) when converted to p will be less than 1000 and would be a 3 or 2 or 1 digit number which would be combination of 1 and 0 so the difference will never be $\geq$ 990 in this case.

$\therefore$ This will never be the answer.

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$\therefore$ Option B. 9 $\leq$ m $\leq$ 13 is the correct answer.