Frequency of a signal sampled at 300 Hz

The analog signal is not printed in the question. This answer assumes the standard textbook signal (Proakis) whose Nyquist rate is exactly 300 Hz:

$$x_a(t) = 3\cos 50\pi t + 10\sin 300\pi t - \cos 100\pi t$$

The same method works for any given $x_a(t)$.

Method

Sampling at $t = nT = n/F_s$ turns an analog frequency $F$ into a discrete-time frequency

$$f = \frac{F}{F_s}\ \text{cycles/sample}, \qquad \omega = 2\pi\frac{F}{F_s}\ \text{rad/sample}$$

Frequencies of each term

Discrete-time signal

$$\begin{aligned} x[n] &= x_a\!\left(\frac{n}{300}\right) \\ &= 3\cos\frac{\pi n}{6} + 10\sin(\pi n) - \cos\frac{\pi n}{3} \end{aligned}$$

Since $\sin(\pi n) = 0$ for every integer $n$, the 150 Hz term gives all-zero samples:

$$x[n] = 3\cos\frac{\pi n}{6} - \cos\frac{\pi n}{3}$$

Answer: the discrete-time frequencies are $f = 1/12,\ 1/2,\ 1/6$ cycles/sample ($\omega = \pi/6,\ \pi,\ \pi/3$ rad/sample). The highest frequency is 150 Hz, so the Nyquist rate is 300 Hz. Sampling at exactly 300 Hz loses the sine term, which shows why $F_s$ must be strictly greater than $2F_{max}$.