Calculating the network function of a filter
You need to consider the system's frequency response

$$H(j\omega)=\frac{\kappa}{j\omega+a}\tag{1}$$

Now you know that $$H(0)=1\tag{2}$$ and $$|H(j\omega_0)|=\frac12\tag{3}$$ (with $\omega_0=10^4$). Note that $H(0)$ is real-valued, whereas $H(j\omega_0)$ will generally be complex-valued. From (1) and (2) you immediately get $\kappa=a$. Furthermore, since the filter is stable we know that $a&gt;0$ (i.e. the pole must lie in the left half-plane of the complex $s$-plane). Combining (3) with (1) gives

$$\frac{a}{\sqrt{\omega_0^2+a^2}}=\frac12\tag{4}$$

from which it should be easy for you to compute $a$.

Note that you don't need the phase value to specify the system. What should be checked is if the phase value is the actual phase value of the system that you just computed. In order to do this you need to verify that

$$\arg\{H(j\omega_0)\}=-\frac{\pi}{3}\tag{5}$$

Luckily that's the case, otherwise the assumption (1) would be wrong. I leave the proof of (5) up to you.