(A ∪ B) = A + B - (A ∩ B)
i.e. (6 ∪ 4) = (multiples of only 6) + (multiples of only 4) - (multiples of both 4 and 6)

Multiples of only 6:
First term = $102$
Last Term = $996$
Number of terms = $\frac{(996 - 102)}{6} + 1 = 150$

Multiples of 4:
First term = $100$
Last Term = $996$
Number of terms = $\frac{(996 - 100)}{4} + 1 = 225$

But, every second number in this series is also divisible by $8$ (with starting and ending number a multiple of 4), Therefore Multiples of only $4 = \frac{225 + 1}{2} = 113$

Multiples of 4 and 6:

Multiples of 4 = 100, 104, 108, 112, .....
Multiples of 6 = 102, 108, 114, 120, ...

First term common = $108$
Last Term common = $996$
Common difference would be $24$

Applying the formula of A.P to find the number of terms:
{NOTE: We are using this formula to find the number of terms above}

$T_n = a + (n - 1)d$
$996 = 108 + (n - 1)24$
$n= 38$

So, $(6 ∪ 4) = (150) + (113) - (38) = 225$

Hence, option B