This is an awesome question!

First, notice that you can pull out a $\frac{1}{3}$ from each of the terms in $\frac{n+1}{3n}$.
For example:

$a_1 = \frac{2}{3} = \frac{1}{3}*2$

$a_2 = \frac{3}{6} = \frac{1}{3}*\frac{3}{2}$

$a_3 = \frac{4}{9} = \frac{1}{3}*\frac{4}{3}$
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The product of the first 53 terms would be:

$a_1*a_2*a_3*.......a_{51}*a_{52}*a_{53}$

In each term, we've pulled out a $\frac{1}{3}$, and since we're multiplying them, this would result in:

$(\frac{1}{3})^{53}$

Now notice what happened when we pulled out the $\frac{1}{3}$ from the sequence in $\frac{n+1}{3n}$.
Each term in the sequence turned into $\frac{n+1}{n}$ (as seen above in the example).

So the product of our sequence looks like this:

$(\frac{1}{3})^{53}$ $* (2*\frac{3}{2}*\frac{4}{3}*\frac{5}{4}.....\frac{53}{52}*\frac{54}{53})$

All of these terms cancel except that last 54, so we're actually left with:

$(\frac{1}{3})^{53}$ $* (54)$

Breaking 54 down:

$(\frac{1}{3})^{53}$ $* (3*3*3*2)$

And this simplifies to:

$(\frac{1}{3})^{50}$ $* (2)$

Or:

$(\frac{2}{3^{50}})$

Giving B as the answer