The temperature is a function of $r,\theta$. The boundary conditions are given for $r=a,b$, so:
$$T(a,\theta)=20 \\ T(b,\theta)=50+10 \cos \theta$$
The angle variable is called $\theta$, not $\phi$ in the problem statement, if case you were confused.
$K$ the thermal conductivity is constant, which means you should be able to write your steady state equation as Laplace equation:
$$\Delta T =\frac{1}{r}\frac{\partial }{\partial r}\left(r \frac{\partial T}{\partial r} \right)+ \frac{1}{r^2} \frac{\partial^2 T}{\partial \theta^2} = 0$$
Turns out you don't actually need the value of thermal conductivity to find the temperature distribution, only to find the heat current later.
Now use separation of variables as intended and the boundary conditions stated above.
Since you appear to have a problem with applying separation of variables to the equation, here's how you do it:
$$\frac{\partial^2 T}{\partial r^2}+\frac{1}{r} \frac{\partial T}{\partial r}+ \frac{1}{r^2} \frac{\partial^2 T}{\partial \theta^2} = 0$$
Substitute $T=f(r)g(\theta)$ into the equation:
$$g \cdot f''+\frac{1}{r} g \cdot f'+ \frac{1}{r^2} f \cdot g'' = 0$$
Divide everything by $f \cdot g$ and multiply by $r^2$:
$$r^2 \frac{f''}{f}+r \frac{f'}{f}+\frac{g''}{g} = 0$$
Now proceed with the separation constant (the sign is chosen for convenience, considering the boundary conditions):
$$r^2 \frac{f''}{f}+r \frac{f'}{f} = \lambda^2$$
$$\frac{g''}{g} = -\lambda^2$$
The above are just ordinary differential equations, the first one gives Bessel functions, the second one trigonometric functions, I hope you know how to write the general solutions and apply the boundary conditions.
Best Answer
The solution to the differential equation is sum of exponential functions.
$$\frac{\partial^2 f}{\partial t^2} - b^2f(t) = 0$$
$$f(x) = k_1 \exp(-bt) +k_2 \exp(bt)$$
If you had opposite sign of $bTX$ then it would be complex conjugate roots and sine-cosine linear combination:
$$\frac{\partial^2 f}{\partial t^2} + b^2f(t) = 0$$
giving
$$f(x) = k_1 \sin(bt) +k_2 \cos(bt)$$