How to prove this result related to divisor function

Question

I need help in deducing this result as a part of question in exercise.

prove that $\sum_{d>Y}\frac{ \tau (d) }{d^2} \ll \frac{ log Y} { Y}$.

I can write LHS as $\sum_{d>Y}\frac{ \tau (d) }{Y^2}\ll \frac{1}{Y^2} \sum_{d>Y}{\tau(d)}$. But I am not able to move forward.