Calculating instantaneous and average power
The instantaneous power is, as you surmised, \$ P= V^2/R=900 \sin^2(\omega t)\,\rm W \$.

The average value of \$ \sin^2(\omega t) \$ over any whole number of cycles is \$ 1/2 \$, so the average power is \$ 450\,\rm W \$.

To address your other formula, \$P=VI\cos\phi\$, observe that the peak voltage is \$150\,\rm V\$, so the RMS voltage is \$150/\sqrt2\,\rm V\$, and the RMS current is \$6/\sqrt 2\,\rm A\$.  Since the load is resistive, the phase angle \$\phi\$ is zero, and the (average) power is \$900/2 = 450\,\rm W\$ as before.