There's 2 ways to approach this question: 1) Try plugging in 54 as a, and work down to check whether y = 5/6x or 2) set up the equations and solve for a. I think Method 1 would be faster, but I'll explain both:

Method 1: Set a = 54
The sequence described can be written as $Term = \frac{1}{3}(Previous Term) + 9$. So if a = 54, then:
$b = \frac{1}{3}(54) + 9 = 27$
$x = \frac{1}{3}(27) + 9 = 18$
$y = \frac{1}{3}(18) + 9 = 15$

Now, we have to check whether y = 5/6x.
$\frac{5}{6}x = \frac{5}{6}(18) = 15 = y$
So, using this method, we know that a must be 54.

Method 2: setting up equations to solve for a
$b = \frac{1}{3}a + 9$
$x = \frac{1}{3}(b) + 9 =\frac{ 1}{3}(\frac{1}{3}a+9) + 9 = \frac{1}{9}a + 12$
$y = \frac{1}{3}(x) + 9 = \frac{1}{3(}\frac{1}{9}a+12) + 9 = \frac{1}{27}a + 13$

We now have x and y in terms of a. Now we plug those into y = 5/6x to solve for a:
$\frac{1}{27}a + 13 = \frac{5}{6}(\frac{1}{9}a+12)$
$5a + 540 = 2a + 702$
$3a = 162$
$a = 54$