Solution:

The answer is (E). The smallest perfect cubes of the form $21 a b^{2}$ and $15 a b$ are such that $a$ and $b$ can only contain the factors 3, 5 and 7 (those contained in 15 or in 21). Let us set $a=3^{l} \cdot 5^{m} \cdot 7^{n}$ and $b=3^{p} \cdot 5^{q} \cdot 7^{r}$ (with non-negative integer exponents), so
$$21 a b^{2} = 3^{l+2p+1} \cdot 5^{m+2q} \cdot 7^{n+2r+1}$$
and
$$15 a b = 3^{l+p+1} \cdot 5^{m+q+1} \cdot 7^{n+r}$$
All the exponents must be multiples of 3, and, in order to obtain the smallest value of $a+b$, they must be as small as possible. So we have the following systems
$$\left\{ \begin{array}{l} l+2p+1=3k_{1} \\ l+p+1=3h_{1} \end{array} \right. \quad ; \left\{ \begin{array}{l} m+2q=3k_{2} \\ m+q+1=3h_{2} \end{array} \right. \quad ; \left\{ \begin{array}{l} n+2r+1=3k_{3} \\ n+r=3h_{3} \end{array} \right.$$
The smallest non-negative integer solutions are $p=0$, $l=2$, $m=1$, $q=1$, $n=1$, $r=2$, so
$$a=3^{2} \cdot 5^{1} \cdot 7^{1}=315,$$
$$b=3^{0} \cdot 5^{1} \cdot 7^{2}=245$$
and
$$a+b=560.$$