Solution:

Given,
- Dhaval has twice as many sisters as he has brothers.
- Deepa has the same number of sisters as she has brothers.

Let's assume,
- Dhaval has x sisters and y brothers.
- Deepa has z sisters and z brothers.

From the given information,
- Dhaval has twice as many sisters as he has brothers, i.e., x = 2y.
- Deepa has the same number of sisters as she has brothers, i.e., z = z.

Now, we need to find out the total number of children Kunal has.

Calculations:

- Let's assume Kunal has n children.
- Out of n children, Dhaval has y+1 siblings (x sisters + y brothers + Dhaval himself).
- And, Deepa has z+1 siblings (z sisters + z brothers + Deepa herself).
- So, the total number of siblings for both Dhaval and Deepa will be (y+1) + (z+1) = n-2 (excluding Dhaval and Deepa themselves).

Now, we have two equations:

- x = 2y
- z = z

We can find out the values of x, y, and z in terms of n:

- x = 2(n-2)/3
- y = (n-2)/3
- z = (n-2)/2

Note: We got the above values by substituting n-2 in place of (y+1) and (z+1) in the equations.

Now, we need to find out the possible values of n for which the above values of x, y, and z are integers.

- We can see that (n-2) should be a multiple of both 2 and 3 for the above values to be integers.
- That is, (n-2) should be a multiple of LCM(2,3) = 6.

Possible values of n:

- If we take n = 8, then (n-2) = 6 is a multiple of 6.
- For n = 9, (n-2) = 7 is not a multiple of 6.
- For n = 10, (n-2) = 8 is a multiple of 6.
- For n = 11, (n-2) = 9 is not a multiple of 6.
- For n = 12, (n-2) = 10 is a multiple of 6.

Therefore, Kunal has 8, 10, and 12 children.

Answer:
- Kunal has 8, 10, or 12 children.