Let $n$ be the total number of students in the six classes that have equal number of students. So $6 \mid n$. In the remaining classes there are $15\%$ more students than in these six classes together, so the number of students in the remaining classes is $0.15n$ more than $n$, i.e., $n + 0.15n = 1.15n$.

The total number of students is $n + 1.15n = 2.15n$.

We are told that $n > 150$ and $2.15n < 400$.

Also, since $0.15n$ must be an integer (number of students), $n$ must be divisible by $20$ (since $0.15n = \frac{3}{20}n$).

So $n$ must be divisible by both $6$ and $20$, i.e., $\operatorname{lcm}(6, 20) = 60$, so $n = 60k$ for some integer $k$.

Now, $n > 150$ and $2.15n < 400$.

The smallest $n$ divisible by $60$ and greater than $150$ is $180$.

Check $n = 180$:

$2.15 \times 180 = 387 < 400$.

The next $n$ is $240$:

$2.15 \times 240 = 516 > 400$.

So only $n = 180$ works.

Therefore, the total number of students in $6$th grade is $387$.