Follow along:
$$\begin{align*} \overline{P}&=f\int_{0}^{\frac{1}{f}} \frac{V_t^2}{R}\:\text{d}t\\\\&=\frac{f}{R}\int_{0}^{\frac{1}{f}} V_t^2\:\text{d}t\\\\&=\frac{f}{R}\left\{\int_{0}^{\frac{0.75}{f}} V^2\:\text{d}t+\int_{0}^{\frac{0.25}{f} } 0\:\text{d}t\right\}\\\\&=\frac{f\,V^2}{R}\int_{0}^{\frac{0.75}{f}} \text{d}t\\\\ &=\frac{f\,V^2}{R}\left[\vphantom{\frac{f\,V^2}{R}\int_{0}^{\frac{0.75}{f}} \text{d}t}t\right]_0^\frac{0.75}{f}\\\\ &=\frac{f\,V^2}{R}\left[\vphantom{\frac{f\,V^2}{R}\int_{0}^{\frac{0.75}{f}} \text{d}t}\frac{0.75}{f}-0\right] \\\\ &=\frac34\frac{V^2}{R}=\frac{\left[\frac{\sqrt{3}}{2}V\right]^2}{R} \end{align*}$$
From this, it's easy to see that \$V_{_\text{RMS}}=\frac{\sqrt{3}}{2}V\$ for a 75% duty cycle. Here, at 75% duty, \$V_{_\text{RMS}}\approx 4.33\:\text{V}\$.
The same process using \$I_t\$ instead of \$V_t\$ would get you to the exact same result.
$$\begin{align*} \overline{P}&=f\int_{0}^{\frac{1}{f}} I_t^2\,R\:\text{d}t\\\\&=f\,R\int_{0}^{\frac{1}{f}} I_t^2\:\text{d}t\\\\ &=f\,R\left\{\int_{0}^{\frac{0.75}{f}} I^2\:\text{d}t+\int_{0}^{\frac{0.25}{f} } 0\:\text{d}t\right\}\\\\ &=f\,I^2\,R\int_{0}^{\frac{0.75}{f}} \text{d}t\\\\ &=f\,I^2\,R\left[\vphantom{\frac{f\,V^2}{R}\int_{0}^{\frac{0.75}{f}} \text{d}t}t\right]_0^\frac{0.75}{f}\\\\ &=f\,I^2\,R\left[\vphantom{\frac{f\,V^2}{R}\int_{0}^{\frac{0.75}{f}} \text{d}t}\frac{0.75}{f}-0\right] \\\\ &=\frac34 \,I^2\,R=\left[\frac{\sqrt{3}}{2}I\right]^2\,R \end{align*}$$
So \$I_{_\text{RMS}}=\frac{\sqrt{3}}{2}\frac{V}{R}\$ for a 75% duty cycle. Here, at 75% duty, \$I_{_\text{RMS}}\approx 43.3\:\text{mA}\$.
Just follow the same process to get the RMS values regardless of repeating waveform.
Take note that \$4.33\:\text{V}\,\cdot\,43.3\:\text{mA}\approx 187.5\:\text{mW}\$, also as expected.