Answer
At $~~t = 2.5\times 10^{-6}~s~~$ the other plate will have maximum positive charge.
Work Step by Step
We can find the period:
$T = \frac{1}{f} = \frac{1}{2.0\times 10^5~Hz} = 5.0\times 10^{-6}~s$
At $~~t = \frac{T}{2}~~$ the other plate will have maximum positive charge.
At $~~t = 2.5\times 10^{-6}~s~~$ the other plate will have maximum positive charge.
