It is worth noting that $R$ in this case denotes the distance from origin to a point $P$ in space. You may be more familiar with $\rho$ instead of $R$.
Here is my attempted solution:
I am assuming that since $0\le\phi\le \pi$ this can be exploited to find a similar range for $R$, as $R=4-3\cos(\phi)$. Doing this I find that $1\le R\le 7$. This thus allows me to set up the following triple integral:
$$\int_0^{2\pi}\int_0^{\pi}\int_1^7R\cdot R^2\sin(\phi)\, \mathrm{d}R\, \mathrm{d}\phi\, \mathrm{d}\theta.$$
However, this gives me an incorrect answer. Can anyone here identify which assumption I have made that is incorrect.
Thanks in advance.
You've set up an integral for a spherical shell, not an elliptical surface. The radius depends on $\phi$, so that needs to be reflected in the limits.
Reverse order of integration from what is usually presented to get
$$2 \pi \int_0^{\pi} d\phi \, \sin{\phi} \, \int_0^{4-3 \cos{\phi}} dR \, R^2$$
which is equal to
$$\frac{2 \pi}{3} \int_0^{\pi} d\phi \, \sin{\phi} \, (4-3 \cos{\phi})^3 = \frac{2 \pi}{3} \int_{-1}^1 du \, (4-3 u)^3$$
The last integral is easily evaluated; I get $6536 \pi$.