Answer
$$\vec{E} = \frac{1}{4\pi\epsilon_0}\pi k{r_0}^2\hat{r}$$
Work Step by Step
Given, $\rho = kr$
Consider a spherical Gaussian surface of radius $r_0$ such that $r_0\le R$
where $R$ is the radius of the sphere.
By Gauss Law,
$$\oint \vec{E}\cdot \vec{da} = \frac{Q_{enc}}{\epsilon_0}$$
Since the charge distribution is radially symmetrical, the direction of the electric field will be radially outwards.
$$\therefore E\times 4\pi {r_0}^2 = \frac{Q_{enc}}{\epsilon_0}$$
$Q_{enc} = \int_{vol.}\rho \times d\tau$
$Q_{enc} = \int_{vol.}kr \times r^2 sin\theta dr d\theta d\phi$
$Q_{enc} = k\int_0^{r_0} r^3 dr \int_0^{\pi} sin\theta d\theta \int_0^{2\pi}d\phi$
$Q_{enc} = k\times \frac{r^4}{4} \times 2 \times 2\pi$
$Q_{enc} = k\pi {r_0}^4$
$$\therefore E\times 4\pi {r_0}^2 = \frac{k\pi{r_0}^4}{\epsilon_0}$$
$$\therefore E = \frac{1}{4\epsilon_0}k{r_0}^2$$
$$\therefore E = \frac{1}{4\pi\epsilon_0}\pi k{r_0}^2$$
$$\therefore \vec{E} = \frac{1}{4\pi\epsilon_0}\pi k{r_0}^2\hat{r}$$
