A king has distributed all his rare jewels in three boxes. The first box contains 1/3 of the rare jewels, while the second box contains k/5 of the rare jewels, for some positive integer value of k. The third box contains 66 rare jewels.
How many rare jewels does the king have?

Let the total number of jewels be X. 
The jewels in the first box would be X/3
The jewels in the second box would be kX/5
And the number of jewels in the third box is 66

The jewels of the second and third boxes should add up to 2X/3 jewels, giving us the relation. 
$$\frac{kX}{5}+66=\frac{2X}{3}$$
$$66=\frac{X\left(10-3k\right)}{15}$$
$$X\left(10-3k\right)=2\times\ 3^2\times\ 5\times\ 11$$

We are given that k is a positive integer. 
Taking k= 1, we get (10 - 3k) to be 7, which is not present on the right-hand side. 
Taking k = 2, we get (10 - 3k) to be 4, which is again not present on the right-hand side. 
Taking k = 3, we get (10 - 3k) to be 1, which is possible on the right-hand side and would give the value fo X to be 990. 

Taking any further value of k would give a negative value of (10 - 3k), which would not be possible. 

Therefore, the king must have had 990 jewels. 

Hence, Option A is the correct answer.