The oscillation frequency in the loop is 10 MHz. If the charge oscillations occur according to the cos law

The oscillation frequency in the loop is 10 MHz. If the charge oscillations occur according to the cos law, then on the capacitor plates q will become equal to half the charge amplitude q (m) after time.

You will need to imagine a “cosine” graph. From which it follows that the maximum charge – q (m) on the capacitor plates will be at the moment of time we need. Then, q will become equal to half the amplitude of the charge q (m) through t = T / 6.
Let’s write the formula for solving the problem:
T = 1 / V;
Let’s substitute:
T = 1/10 * 10 ^ 6 Hz;
We get:
T = 0.1 * 10 ^ -6 s.
Find t:
t = 0.1 * 10 ^ -6 / 6;
t = 0.017 * 10 ^ -6 s.
Answer: t = 0.017 * 10 ^ -6 s.