A string 0.90 m long is fixed at both ends and vibrates in its third harmonic…

Step 1. The frequency. $f_3 = \dfrac{3 \times 60}{2 \times 0.90} = 100$ Hz.

Step 2. The wavelength. $\lambda = \dfrac{v}{f} = \dfrac{60}{100} = 0.60$ m.

Step 3. The nodes. They are $\dfrac{\lambda}{2} = 0.30$ m apart: at $0$, $0.30$, $0.60$ and $0.90$ m.

Step 4. Count. There are $4$ nodes. In general, harmonic $n$ has $n + 1$ nodes.