The given problem is related to the ratio of powers present in the 3rd harmonic and 5th harmonic of a square periodic signal. Let's understand the solution step by step:

1. Waveform of a square periodic signal: A square periodic signal can be represented as a series of odd harmonics. The waveform is characterized by square-shaped pulses that repeat at regular intervals. The fundamental frequency is denoted as the 1st harmonic.

2. Harmonics in a square periodic signal: Harmonics are integer multiples of the fundamental frequency. In a square periodic signal, the amplitude of odd harmonics decreases as the harmonic number increases. The 3rd harmonic and 5th harmonic are two such odd harmonics.

3. Power in a signal: Power is directly proportional to the square of the amplitude of a signal. Therefore, to compare the powers present in the 3rd harmonic and 5th harmonic, we need to compare the squares of their amplitudes.

4. Ratio of powers: Let's assume the amplitude of the 3rd harmonic is A3 and the amplitude of the 5th harmonic is A5. The ratio of powers can be calculated as (A3^2)/(A5^2).

5. Relationship between harmonic amplitudes: The amplitudes of the harmonics in a square periodic signal follow a specific relationship. The amplitude of the nth harmonic is given by 1/n times the amplitude of the fundamental frequency. Therefore, we can write A3 = (1/3)A1 and A5 = (1/5)A1, where A1 is the amplitude of the fundamental frequency.

6. Calculating the ratio of powers: Substituting the values of A3 and A5 in the ratio of powers equation, we get (A3^2)/(A5^2) = [(1/3)A1]^2 / [(1/5)A1]^2 = (1/9) / (1/25) = 25/9.

Hence, the correct answer is option 'C' - 25/9.

Note: The solution assumes that the square periodic signal is an ideal square wave with perfect odd harmonics. In practical scenarios, there might be some distortion and additional harmonics present, which can affect the exact ratio of powers.