3.3 V through a 2 m cable : can it work?
The voltage at the second PCB can be calculated through

$$ V_{2nd,PCB}  = V_{3V3} - 2 \cdot I_{2nd,PCB} \cdot \frac{l_{cable}\cdot \rho}{A_{cable}}  $$

Where $$ \rho  \approx 1.7 \cdot 10^{-8} \Omega m$$ 

If your second PCB's lower supply voltage threshold is alright with this voltage drop, you should not worry about it. Just make sure to add some additional capacitance at the input of the second PCB in order to cope with high transients (The cabble adds additional inductance).