$$f\left(a\right)=a^2+3a\ ----eqn\ \left(1\right)$$
$$f\left(b\right)=b^2+3b\ ----eqn\ \left(2\right)$$
eqn (1) - eqn (2) ==> f(a-b)
$$f\left(a-b\right)=\left(a^2+3a\right)-\left(b^2+3b\right)$$
$$=a^2+3a-b^2-3b$$
$$=\left(a^2-b^2\right)+3\left(a-b\right)$$
$$Therefore,\ 3f\left(a-b\right)=3\left[\left(a^2-b^2\right)+3\left(a-b\right)\right]$$
$$=3a^2+9a-3b^2-9b$$
$$=3\left(a^2-b^2\right)+9\left(a-b\right)$$
Question: The value of 3f(a-b) is how many times greater than b?
So, find the values of both 'a' and 'b' to if 3f(a-b)>b?
Statement 1: b=3
Since there is no information about the value of a, hence, statement 1 is NOT SUFFICIENT.

Statement 2: a is a positive integer
This means that 'a' can be any value between 1 and infinite. Since there is no definite value for 'b' in this statement, then statement 2 is also NOT SUFFICIENT.

Combine both statements together:
From statement 1: b=3
From statement 2: 'a' is a positive integer
If a=1 and b=3
$$=3\left(a^2-b^2\right)+9\left(a-b\right)$$
$$=3\left(1^2-3^2\right)+9\left(1-3\right)$$
3f(a - b) = (3 - 27) + (9 - 27)
= -42
Therefore, 3f(a - b) < b

If a=10 and b=3
$$=3\left(a^2-b^2\right)+9\left(a-b\right)$$
$$=3\left(10^2-3^2\right)+9\left(10-3\right)$$
3f(a - b) = (300 - 27) + (90 - 27)
= 336
Therefore, 3f(a - b) > b
Due to the udefined value of 'a', it is not possible to determine if 3f(a - b) > b or 3f(a - b) < b. Hence, both statements combined ARE NOT SUFFICIENT.

Answer = option E

Thanks