I'll provide a different approach than Jan did (he shouldn't have computed "RMS" for power), which may be a little more directly understandable. I get Jan's average power value, though. (I see he just deleted his posted answer.)
Assume \$t_1=5\:\mu\text{s}\$, \$t_2=7\:\mu\text{s}\$, and \$t_3=13\:\mu\text{s}\$ (\$t_0=0\:\text{s}\$). We also know these details:
$$\begin{align*}
V_t\:\,\bigg|_{t_0}^{t_1}&=\left(6 \,\frac{t}{t_1}-4\right)\:\text{V}\\\\
V_t\:\,\bigg|_{t_1}^{t_2}&=2\:\text{V}\\\\
V_t\:\,\bigg|_{t_2}^{t_3}&=-4\:\text{V}\\\\
\end{align*}$$
The average power is integrated energy over the time period, divided by the time period itself. (Average power, as opposed to instantaneous power, is the finite work done, divided by the finite time over which that work was performed.) So each of the following individual integrals yield energy during their period of time. Dividing that total energy by the time period will provide the power over that time period:
$$\begin{align*}
\overline{P}\:\,\bigg|_{t_0}^{t_3}&=\frac{1}{t_3-t_0}\left[\int_{t_0}^{t_1}\frac{V_t^2}{R}\:\text{d}t+\int_{t_1}^{t_2}\frac{V_t^2}{R}\:\text{d}t+\int_{t_2}^{t_3}\frac{V_t^2}{R}\:\text{d}t\right]\\\\
&=\frac{1}{t_3-t_0}\frac1{R}\left[\int_{t_0}^{t_1}V_t^2\:\text{d}t+\int_{t_1}^{t_2}V_t^2\:\text{d}t+\int_{t_2}^{t_3}V_t^2\:\text{d}t\right]\\\\
&\text{ignoring units for now and substituting in }t_0=0\:\text{s},\\\\
&=\frac{1}{R\,t_3}\left[\int_{0}^{t_1}\left(6 \cdot \frac{t}{t_1}-4\right)^2\:\text{d}t+\int_{t_1}^{t_2}\left(2\right)^2\:\text{d}t+\int_{t_2}^{t_3}\left(-4\right)^2\:\text{d}t\right]\\\\
&=\frac{1}{R\,t_3}\left[\int_{0}^{t_1}\left(6 \cdot \frac{t}{t_1}-4\right)^2\:\text{d}t+4\,\left(t_2-t_1\right)+16\,\left(t_3-t_2\right)\right]\\\\
&=\frac{1}{R\,t_3}\left\{\left[\left(12\, \frac{t^2}{t_1^2}-24\, \frac{t}{t_1}+16\right)\cdot t\right]\bigg|_{0}^{t_1}+4\,\left(t_2-t_1\right)+16\,\left(t_3-t_2\right)\right\}\\\\
&=\frac{1}{R\,t_3}\bigg\{4\, t_1+4\,\left(t_2-t_1\right)+16\,\left(t_3-t_2\right)\bigg\}\\\\
&=\frac{1}{R\,t_3}\bigg\{16\,t_3-12\,t_2\bigg\}\\\\
&\text{putting units back in,}\\\\
&=\frac{16\:\text{V}^2\cdot 13\:\mu\text{s}-12\:\text{V}^2\cdot 7\:\mu\text{s}}{75\:\Omega\cdot 13\:\mu\text{s}}\approx 127.18\:\text{mW}
\end{align*}$$
Note that this disagrees with the answer you provide. So perhaps you may wish to reconsider it. (Or someone will point out my error, perhaps?)
I decided to do an LTspice run, just to be sure of the above results. (It never hurts.) Here is the result:
It appears that LTspice arrives at the same value. I think that's a good thing.
