The general solution to $\Delta u\equiv0$ on $x^2+y^2\leq 1$ is known to be $$u(r,\theta)=\frac{\alpha_0}{2}+\sum_{n=1}^{\infty} r^n\left(\alpha_n \cos(n\theta)+\beta_n \sin(n\theta)\right)$$
Here $r\in [0,1]$ and $\theta\in[-\pi,\pi)$. Applying $u_r(1,\theta)=3\sin(2\theta)+1$ yields $$3\sin(2\theta)+1=\sum_{n=1}^{\infty}\left((n\alpha_n)\cos(n\theta)+(n\beta_n)\sin(n\theta)\right)$$ If we multiply both sides by $d\theta$ and integrate from $\theta=-\pi$ to $\theta=\pi$ we get $1=0$ which is absurd.
Do you think there is a typo in this problem, or is there a flaw with my logic?