Electric field of given charge density

First of all, it helps to imagine the geometry of your problem. So to imagine the physical form of $$ \rho(\vec x) = \rho_0 \delta(x_1) \delta(x_2) \quad,$$ which is just an infinite line along the $x_3$-axis. So maybe this helps you to calculate the electric field, since now you are able to see the symmetry of the problem and you could know that cylindrical coordinates and Gauss' law might help you out a little.

Edit: Since your question seems to ask how to calculate the given Integral not how to calculate the electric field i maybe did not answer you question with whats written above so:

The integral $$ \int_{\mathbb R^3} \text d ^3x' \frac{\rho_0 \delta(x_1') \delta(x_2')}{|\vec x - \vec x'|^3}(\vec x - \vec x') $$ is a set of $3$ integrals. Lookin at the first component: \begin{align} 4 \pi \epsilon_0 E_1 &= \int_{\mathbb R^3} \text d ^3x' \frac{\rho_0 \delta(x'_1) \delta(x'_2)}{((x_1 - x_1')^2 + (x_2 - x_2')^2 + (x_3 - x_3')^2)^{\frac 3 2}}(x_1 - x'_1)\\ &=\int_{\mathbb R}\text{d}x'_3 \frac{\rho_0 x_1}{(x_1^2 + x_2^2 + (x_3 - x_3')^2)^{\frac 3 2}}\\ &= 2 \rho_0 \frac{x_1}{x_1^2 + x_2^2} \end{align} The same goes for the second component: \begin{align} 4 \pi \epsilon_o E_2 = 2\rho_0 \frac{x_2}{x_1^2 + x_2^2} \end{align} The third component is $0$ since: $$ \int_{\mathbb R}\text d x'_3 \frac{(x_3 - x'_3)}{(x_1^2 + x_2^2 + (x_3 - x_3')^2)^{\frac 3 2}} = 0 $$ I leave it to you to interpret the result.

Actually, $$\int\delta(x-x')f(x')dx'=f(x)~.$$ By substituting $$f(x)=\frac{\vec{x}-\vec{x'}}{|\vec{x}-\vec{x}'|^3}$$ you will get the desired result.