Answer
$\int_{vol.} (\vec{\triangledown}\cdot \vec{v})d\tau = \oint_S \vec{v}\cdot\vec{da} = \frac{1}{4}\pi R^4$
Work Step by Step
Let's consider the octant where
$r$ varies from $0$ to $R$
$\theta$ varies from $0$ to $\pi\over 2$
$\phi$ varies from $0$ to $\pi\over 2$
Given $\vec{v} = r^2cos\theta \hat{r} + r^2 cos\phi\hat{\theta} - r^2 cos\theta sin\phi \hat{\phi} $
$\vec{\triangledown}\cdot \vec{v} = \frac{1}{r^2 sin\theta}\big[\frac{\partial}{\partial r}(r^2sin\theta \cdot r^2 cos\theta) + \frac{\partial}{\partial \theta}(rsin\theta \cdot r^2 cos\phi) + \frac{\partial}{\partial \phi} (r\cdot (-r^2 cos\theta sin\phi))\big]$
$\vec{\triangledown}\cdot \vec{v} = \frac{1}{r^2 sin\theta}\big[ sin\theta cos\theta 4r^3+ r^3cos\phi cos\theta -r^3 cos\theta cos\phi\big]$
$\vec{\triangledown}\cdot \vec{v} = 4rcos\theta$
$\int_{vol.}(\vec{\triangledown}\cdot \vec{v}) d\tau = \int_{vol.} 4rcos\theta \cdot r^2 sin\theta dr d\theta d\phi $
$\int_{vol.}(\vec{\triangledown}\cdot \vec{v}) d\tau = 4\times \int_0^R r^3 dr\times \int_0^{\pi \over 2} cos\theta sin\theta d\theta \times \int_0^{\pi \over 2} d\phi$
$\int_{vol.}(\vec{\triangledown}\cdot \vec{v}) d\tau = 4\times \frac{R^4}{4}\times \frac{1}{2} \times \frac{\pi}{2} = \frac{1}{4} \pi R^4$
Divide the surface into 4 parts where
$S_1$ represents the part on the on the $xy$ plane. Hence $\theta = \pi/2$
$\int_{S_1} \vec{v}\cdot \vec{da} = \int_{S_1} (r^2 cos\phi \hat{\theta})\cdot (r dr d\phi \hat{\theta})$
$\int_{S_1} \vec{v}\cdot \vec{da} = \int_0^R r^3 dr \int_0^{\pi/2} cos\phi d\phi$
$\int_{S_1} \vec{v}\cdot \vec{da} = \frac{R^4}{4} \times 1 = \frac{1}{4} R^4$
$S_2$ represents the part on the on the $yz$ plane. Hence $\phi = \pi/2$
$\int_{S_2} \vec{v}\cdot \vec{da} = \int_{S_2} (r^2 cos\theta \hat{r} - r^2 cos\theta \hat{\phi})\cdot (r dr d\theta \hat{\phi})$
$\int_{S_2} \vec{v}\cdot \vec{da} = \int_{S_2} - r^3 cos\theta dr d\theta $
$\int_{S_2} \vec{v}\cdot \vec{da} = -\int_0^R r^3 dr \int_0^{\pi/2} cos\theta d\theta$
$\int_{S_2} \vec{v}\cdot \vec{da} = -\frac{R^4}{4} \times 1 = -\frac{1}{4} R^4$
$S_3$ represents the part on the on the $xz$ plane. Hence $\phi = 0$
$\int_{S_3} \vec{v}\cdot \vec{da} = \int_{S_3} (r^2 cos\theta \hat{r} + r^2 cos\phi \hat{\theta})\cdot (-r dr d\theta \hat{\phi}) = 0$
$S_4$ represents the curved part of the sphere . Hence $r = R$
$\int_{S_4} \vec{v}\cdot \vec{da} = \int_{S_4} (R^2 cos\theta \hat{r} + R^2 cos\phi \hat{\theta}-R^2 cos\theta sin\phi \hat{\phi})\cdot (R^2 sin\theta d\theta d\phi \hat{r})$
$\int_{S_4} \vec{v}\cdot \vec{da} = \int_{S_4} (R^4 cos\theta sin\theta d\theta d\phi) $
$\int_{S_4} \vec{v}\cdot \vec{da} = R^4\int_0^{\pi /2} cos\theta sin\theta d\theta \int_0^{\pi/2} d\phi$
$\int_{S_4} \vec{v}\cdot \vec{da} = R^4\times \frac{1}{2} \times \frac{\pi}{2} = \frac{1}{4}\pi R^4$
$\oint_S \vec{v}\cdot \vec{da}= \int_{S_1}\vec{v}\cdot \vec{da}+\int_{S_2}\vec{v}\cdot \vec{da}+\int_{S_3}\vec{v}\cdot \vec{da}+\int_{S_4}\vec{v}\cdot \vec{da}$
$\oint_S \vec{v}\cdot \vec{da}= \frac{1}{4}\pi R^4$
Hence
$\int_{vol.} (\vec{\triangledown}\cdot \vec{v})d\tau = \oint_S \vec{v}\cdot\vec{da}$
Hence divergence theorem is verified
