Answer
For $r\ge R$, $\vec{E} = {\sigma\times R^2 \over \epsilon_0\times r^2}\hat{r}$
For $r< R, \vec{E}=0$
Work Step by Step
Consider a gaussian spherical surface of radius $r$ where $r\ge R$
According to Gauss Law,
$$\oint \vec{E}\cdot \vec{da} = \frac{Q_{enc}}{\epsilon_{0}} = \frac{\sigma\times 4\pi R^2}{\epsilon_0}$$
At all places of the Gaussian surface, $\vec{E}$ and $\vec{da}$ point in the same direction i.e $\hat{r}$
$$E\times 4\pi r^2 = {\sigma\times 4\pi R^2 \over \epsilon_0}$$
$$\vec{E} = {\sigma\times R^2 \over \epsilon_0\times r^2}\hat{r}$$
Now, For $r< R, Q_{enc} = 0$ and hence $\vec{E}=0$
