As always, the first thing you need to do is calculate your properties.
\begin{align}
A_A &= 150\cdot10 + 6\cdot(200-10) = 2640\text{ mm}^2 \\
\overline{y}_A &= \dfrac{150\cdot10\cdot195 + 6\cdot190\cdot95}{2640} = 151.8\text{ mm} \\
I_A &= \dfrac{150\cdot10^3}{12}+150\cdot10\cdot(151.8-195)^2+\dfrac{6\cdot190^3}{12}+6\cdot190\cdot(151.8-95)^2 \\
I_A &=9919274\text{ mm}^4 \\
A_B &= 150\cdot10 + 6\cdot(100-10) = 2040\text{ mm}^2 \\
\overline{y}_B &= \dfrac{150\cdot10\cdot95 + 6\cdot90\cdot45}{2040} = 81.76\text{ mm} \\
I_B &= \dfrac{150\cdot10^3}{12}+150\cdot10\cdot(81.76-95)^2+\dfrac{6\cdot90^3}{12}+6\cdot90\cdot(81.76-45)^2 \\
I_B &=1369647\text{ mm}^4
\end{align}
Now, you need to observe the three sources of stress in your structure:
- The axial load at point B;
- The moment due to the eccentricity of the load at point B;
- The moment due to the load $q$ to be determined.
So, what you need to do is calculate the stresses due to the known sources (#1 and #2) to see what slack you have for the load $q$. Strictly speaking you should check points A and B, but I'll only do A here.
So, the axial load at point B is equal to $70\cdot2040=142800\text{ N}$. This immediately gives us a uniform stress of
$$\sigma_1 = \dfrac{142800}{2640} = 54.09\text{ N/mm}^2$$
However, the resultant force of the load at point B is located at the centroid of the section at point B, which is different from the centroid at point A. This causes a moment, which is equal to
\begin{align}
M &= 142800\left((200-151.8)-(100-81.76)\right) = 4278288\text{ Nmm} \\
\therefore \sigma_2 &= \dfrac{My}{I} = \dfrac{4278288\cdot(200-151.8)}{9919274} =20.79\text{ N/mm}^2
\end{align}
Note the stress is being calculated in the top fiber, since that is where the tensile stresses occur.
So, just these loads give you already $\sigma = 54.09+20.79=74.88\text{ N/mm}^2$, meaning you are allowed an increment of $\sigma_3 = 115-74.88 = 40.12\text{ N/mm}^2$.
You are correct that you need to back-calculate the load via stress, so:
\begin{align}
M &= \dfrac{qL^2}{2} \\
\sigma_3 &= \dfrac{My}{I} \\
\therefore \sigma_3 &= \dfrac{qL^2y}{2I} \\
\therefore q &= \dfrac{2I\sigma_3}{yL^2} = \dfrac{2\cdot9919274\cdot40.12}{(200-151.8)\cdot2500^2} = 2.64\text{ N/mm}
\end{align}
I may have bundled the calculations, but that's the general gist of it.
